Answer

(c) Standard electrode potential of copper electrode can be calculated by constructing a concentration cell composed of two half cell reactions in which concentration of species on left hand and right hand side are unity. In such case cell potential is equal to standard electrode potential.

$$ \underbrace{Pt(s)|H_2(g, 1 bar)| \mid H^+(aq, 1 M)}_{\text {Oxidation half cell reaction }} \mid \underbrace{|Cu^{2+}(aq, 1 M)| Cu} _{\text {Reduction half cell reaction }} $$

• Option (a): The hydrogen gas pressure is 0.1 bar instead of the standard 1 bar, which means it does not meet the standard conditions required for measuring the standard electrode potential.

• Option (b): The concentration of ( \text{Cu}^{2+} ) is 2 M instead of the standard 1 M, which means it does not meet the standard conditions required for measuring the standard electrode potential.

• Option (d): Both the hydrogen gas pressure is 0.1 bar and the concentration of ( \text{H}^+ ) is 0.1 M, neither of which meet the standard conditions required for measuring the standard electrode potential.

Thinking Process

This problem includes concept of Nernst equation and its transformation to equation of straight line.

Answer

Electrode potential for Mg electrode varies according to the equation

$$ \begin{aligned} & E_{Mg^{2+} / Mg}=E_{Mg^{2+} / Mg}^{\circ}-\frac{0.059}{2} \log \frac{1}{[Mg^{2+}]} \\ & E_{Mg^{2+} / Mg}=E_{Mg^{2+} / Mg}^{\circ}+\frac{0.059}{2} \log [Mg^{2+}] \\ & E_{Mg^{2+} / Mg}=\frac{0.059}{2} \log [Mg^{2+}]+E_{Mg^{2+} / Mg}^{\circ} \end{aligned} $$

This equation represents equation of straight line. It can be correlated as

$ \begin{array}{cccc} E_{\mathrm{Mg}^{2+} / \mathrm{Mg}} & =\left(\frac{0.059}{2}\right) & \log \left[\mathrm{Mg}^{2+}\right] & +E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ} \\ \uparrow & \uparrow & \uparrow & \uparrow \\ \text{ Y }& \text { M } & \text{ X } & \text{ +C } \end{array} $

So, intercept $(C)=E_{Mg^{2+}}^{\circ} Mg$

Thus, equation can be diagramatically represented as

1. Option 1:

   • The graph in Option 1 shows a negative slope, which implies that the electrode potential decreases as the concentration of ( \text{Mg}^{2+} ) increases. However, according to the given equation ( E_{Mg^{2+} / Mg} = E_{Mg^{2+} / Mg}^{\circ} + \frac{0.059}{2} \log [Mg^{2+}] ), the slope should be positive (( \frac{0.059}{2} )), indicating that the electrode potential increases with an increase in ( \text{Mg}^{2+} ) concentration.

2. Option 2:

   • The graph in Option 2 shows a horizontal line, which implies that the electrode potential is constant and does not change with the concentration of ( \text{Mg}^{2+} ). This contradicts the given equation, which indicates a linear relationship between ( E_{Mg^{2+} / Mg} ) and ( \log [Mg^{2+}] ) with a positive slope.

3. Option 3:

   • The graph in Option 3 shows a curve, which implies a non-linear relationship between the electrode potential and the concentration of ( \text{Mg}^{2+} ). However, the given equation ( E_{Mg^{2+} / Mg} = E_{Mg^{2+} / Mg}^{\circ} + \frac{0.059}{2} \log [Mg^{2+}] ) represents a linear relationship, not a curved one. Therefore, this option is incorrect as it does not represent a straight line.

Thinking Process

This problem is based on thermodynamical concept of intensive and extensive property. During answering this question must keep in mind that intensive property is independent on number of particles and extensive property is dependent on number of particles.

Answer

(c) $E_{\text {cell }}$ is an intensive property as it does not depend upon mass of species (number of particles) but $\Delta_{r} G$ of the cell reaction is an extensive property because this depends upon mass of species (number of particles).

• (a) $E_{\text {cell }}$ and $\Delta_{r} G$ of cell reaction both are extensive properties: This is incorrect because $E_{\text {cell }}$ (cell potential) is an intensive property, meaning it does not depend on the amount of substance or the size of the system. In contrast, $\Delta_{r} G$ (Gibbs free energy change) is an extensive property, meaning it does depend on the amount of substance.

• (b) $E_{\text {cell }}$ and $\Delta_{r} G$ of cell reaction both are intensive properties: This is incorrect because while $E_{\text {cell }}$ is indeed an intensive property, $\Delta_{r} G$ is an extensive property. Intensive properties do not depend on the size or amount of material in the system, whereas extensive properties do.

• (d) $E_{\text {cell }}$ is an extensive property while $\Delta_{r} G$ of cell reaction is an intensive property: This is incorrect because $E_{\text {cell }}$ is an intensive property, not an extensive one. Conversely, $\Delta_{r} G$ is an extensive property, not an intensive one.

Answer

(b) Cell emf The difference between the electrode potential of two electrodes when no current is drawn through the cell is called cell emf.

• (a) Cell potential: This term is often used interchangeably with cell emf, but it can also refer to the potential difference under conditions where current is flowing, which is not the case in the given scenario.

• (c) Potential difference: This is a general term that refers to the difference in electric potential between two points. It does not specifically refer to the condition where no current is drawn through the cell.

• (d) Cell voltage: This term typically refers to the voltage across the cell when it is operating and current is flowing. It does not specifically describe the condition where no current is drawn through the cell.

Answer

(d) An inert electrode in a cell provide surface for either oxidation or for reduction reaction by conduction of electrons through its surface but does not participate in the cell reaction.

It does not provide surface for redox reaction.

• (a) It does not participate in the cell reaction: This statement is correct. An inert electrode does not participate in the chemical reaction itself; it only provides a surface for the reaction to occur.

• (b) It provides surface either for oxidation or for reduction reaction: This statement is correct. An inert electrode provides a surface for either the oxidation or reduction reaction to take place.

• (c) It provides surface for conduction of electrons: This statement is correct. An inert electrode conducts electrons, allowing the redox reactions to occur at its surface.

Answer

(c) If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value $1.1 V$.

At this stage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in opposite direction.

Hence, this works as an electrolytic cell.

• (a) ( E_{\text{cell}} = 0 ): When the cell potential is zero, it means that the cell is at equilibrium and no net reaction occurs. This does not cause the cell to behave like an electrolytic cell, as there is no driving force for the reaction to proceed in either direction.

• (b) ( E_{\text{cell}} > E_{\text{ext}} ): In this case, the cell potential is greater than the external potential, which means the galvanic cell will continue to operate normally, driving the spontaneous reaction forward. It will not behave like an electrolytic cell because the external potential is not sufficient to reverse the reaction.

• (d) ( E_{\text{cell}} = E_{\text{ext}} ): When the cell potential is equal to the external potential, the system is in a state of equilibrium where no net current flows. This does not cause the cell to behave like an electrolytic cell, as there is no external force driving the reaction in the reverse direction.

Answer

(c) Solution consists of electrolytes is known as electrolytic solution and conductivity of electrolytic solution depends upon the following factors

(i) Size of ions As ion size increases, ion mobility decreases and conductivity decreases.

(ii) Viscosity of solution Greater the viscosity of the solvent lesser will be the conductivity of the solution.

(iii) Solvation of ions Greater the solvation of ions of an electrolyte lesser will be the electrical conductivity of the solution.

(iv) Temperature of medium Conductivity of solution increases with increase in temperature.

• (a) Conductivity of solution depends upon size of ions: This statement is correct because as the size of ions increases, their mobility decreases, leading to a decrease in conductivity.

• (b) Conductivity depends upon viscosity of solution: This statement is correct because greater viscosity of the solvent results in lower ion mobility, thereby reducing the conductivity of the solution.

• (d) Conductivity of solution increases with temperature: This statement is correct because increasing the temperature generally increases the mobility of ions, which in turn increases the conductivity of the solution.

Thinking Process

This problem includes concept of electrochemical series and standard reduction potential of the metal.

Higher the negative value of standard reduction potential, strongest will be the reducing agent.

As value of SRP increases from negative to positive value nature of species changes from reducing to oxidising nature.

Answer

(b) Here, out of given four options standard reduction potential of chromium has highest negative value hence most powerful reducing agent is chromium.

• (a) Cl⁻: Chloride ion (Cl⁻) has a standard reduction potential of 1.36 V for the Cl₂/Cl⁻ couple, which is a positive value. This indicates that Cl⁻ is more likely to gain electrons and be reduced rather than lose electrons and act as a reducing agent. Therefore, it is not the strongest reducing agent.

• (c) Cr³⁺: The standard reduction potential for the Cr³⁺/Cr couple is -0.74 V. While this is a negative value, indicating that Cr³⁺ can act as a reducing agent, it is not as negative as the reduction potential for Cr (which is -0.74 V for Cr³⁺/Cr). Therefore, Cr³⁺ is not the strongest reducing agent compared to Cr.

• (d) Mn²⁺: The standard reduction potential for the MnO₄⁻/Mn²⁺ couple is 1.51 V, which is a positive value. This indicates that Mn²⁺ is more likely to gain electrons and be reduced rather than lose electrons and act as a reducing agent. Therefore, it is not the strongest reducing agent.

Answer

(c) Higher the positive value of standard reduction potential of metal ion higher will be its oxidising capacity.

Since, $E_MnO_4^{\circ} / Mn^{2+}$ has value equal to $1.51 V$ hence it is the strongest oxidising agent.

• (a) $Cl^{-}$: The standard reduction potential of $Cl_2/Cl^{-}$ is lower than that of $MnO_4^{-}/Mn^{2+}$, indicating that $Cl^{-}$ is a weaker oxidising agent compared to $MnO_4^{-}$.

• (b) $Mn^{2+}$: The standard reduction potential of $Mn^{2+}/Mn$ is lower than that of $MnO_4^{-}/Mn^{2+}$, indicating that $Mn^{2+}$ is a weaker oxidising agent compared to $MnO_4^{-}$.

• (d) $Cr^{3+}$: The standard reduction potential of $Cr^{3+}/Cr$ is lower than that of $MnO_4^{-}/Mn^{2+}$, indicating that $Cr^{3+}$ is a weaker oxidising agent compared to $MnO_4^{-}$.

Answer

(b)

On moving top to bottom SRP value decreases from positive to negative value which will increase the reducing capacity. So, the correct option is (b).

• Option (a): $Cr^{3+}<Cl^{-}<Mn^{2+}<Cr$

  • Incorrect because $Mn^{2+}$ has a higher SRP value (1.51 V) than $Cl^{-}$ (1.36 V), so $Mn^{2+}$ should come before $Cl^{-}$ in the order of reducing power.

• Option (c): $Cr^{3+}<Cl^{-}<Cr_2 O_7^{2-}<MnO_4^{-}$

  • Incorrect because $Cr_2 O_7^{2-}$ and $MnO_4^{-}$ are not included in the given SRP values table, making it impossible to determine their correct order relative to $Cr^{3+}$ and $Cl^{-}$.

• Option (d): $Mn^{2+}<Cr^{3+}<Cl^{-}<Cr$

  • Incorrect because $Mn^{2+}$ has a higher SRP value (1.51 V) than $Cr^{3+}$ (1.33 V), so $Mn^{2+}$ should come after $Cr^{3+}$ in the order of reducing power.

Answer

(d) $E_{MnO_4-M Mn^{2+}}^{o}$ has + ve value equal to $1.51 V$ which is highest among given four choices. $So, Mn^{2+}$ is most stable ion in its reduced form.

• (a) $Cl^{-}$: The standard reduction potential for $Cl_2/Cl^{-}$ is +1.36 V, which is lower than the +1.51 V for $MnO_4^-/Mn^{2+}$. Therefore, $Cl^{-}$ is less stable in its reduced form compared to $Mn^{2+}$.

• (b) $Cr^{3+}$: The standard reduction potential for $Cr_2O_7^{2-}/Cr^{3+}$ is +1.33 V, which is also lower than the +1.51 V for $MnO_4^-/Mn^{2+}$. Hence, $Cr^{3+}$ is less stable in its reduced form compared to $Mn^{2+}$.

• (c) $Cr$: The standard reduction potential for $Cr^{3+}/Cr$ is -0.74 V, which is significantly lower than the +1.51 V for $MnO_4^-/Mn^{2+}$. This makes $Cr$ much less stable in its reduced form compared to $Mn^{2+}$.

Answer

(a) $E_{Cr^{3+} / Cr}^{\circ}$ has most - ve value equal to -0.74 among given four choices. $So, Cr^{3+}$ is the most stable oxidised species.

• (b) $MnO_4^{-}$: The $E_{MnO_4^{-} / Mn^{2+}}^{\circ}$ value is highly positive, indicating that $MnO_4^{-}$ is a strong oxidizing agent and not a stable oxidized species.

• (c) $Cr_2O_7^{2-}$: The $E_{Cr_2O_7^{2-} / Cr^{3+}}^{\circ}$ value is also positive, suggesting that $Cr_2O_7^{2-}$ is a strong oxidizing agent and not a stable oxidized species.

• (d) $Mn^{2+}$: The $E_{Mn^{2+} / Mn}^{\circ}$ value is less negative than $E_{Cr^{3+} / Cr}^{\circ}$, indicating that $Mn^{2+}$ is less stable as an oxidized species compared to $Cr^{3+}$.

Answer

(c) The quantity of charge required to obtain one mole of aluminium from $Al_2 O_3$ is equal to number of electron required to convert $Al_2 O_3$ to $Al$.

$$ Al^{3+}(aq) \stackrel{+3 e}{\longrightarrow} Al(s) $$

Hence, total $3 F$ is required.

• Option (a) $1 F$: This is incorrect because $1 F$ (Faraday) corresponds to the charge required to deposit or liberate one mole of a monovalent ion. Since aluminium is trivalent ($Al^{3+}$), it requires 3 moles of electrons (3 Faradays) to reduce one mole of $Al^{3+}$ to aluminium metal.

• Option (b) $6 F$: This is incorrect because it overestimates the charge required. For one mole of aluminium, only 3 Faradays are needed, as each aluminium ion requires 3 electrons for reduction. Therefore, 6 Faradays would be double the necessary amount.

• Option (d) $2 F$: This is incorrect because it underestimates the charge required. Since each $Al^{3+}$ ion needs 3 electrons to be reduced to aluminium metal, 2 Faradays would not be sufficient to reduce one mole of $Al^{3+}$ ions.

Answer

(d) Cell constant is defined as the ratio of length of object and area of cross section.

$$ G=\frac{l}{A} $$

Since, $l$ and $A$ remain constant for any particular object hence value of cell constant always remains constant.

• (a) The cell constant does not change with the change of electrolyte because it is a geometric property of the cell, dependent only on the physical dimensions (length and area) of the cell, not on the type of electrolyte used.

• (b) The cell constant does not change with the change of concentration of electrolyte because it is independent of the properties of the solution. It is solely determined by the physical dimensions of the cell.

• (c) The cell constant does not change with the temperature of the electrolyte because temperature affects the conductivity of the electrolyte, not the physical dimensions of the cell. The cell constant is a fixed value based on the cell’s geometry.

Answer

(a) While charging the lead storage battery the reaction occurring on cell is reversed and $PbSO_4(s)$ on anode and cathode is converted into $Pb$ and $PbO_2$ respectively

Hence, option (a) is the correct choice

The electrode reactions are as follows

At cathode $PbSO_4(s)+2 e^{-} \rightarrow Pb(s)+SO_4{ }^{2-}(aq)$ (Reduction)

At anode $PbSO_4(s)+2 H_2 O \rightarrow PbO_2(s)+SO_4{ }^{2-}+4 H^{+}+2 e^{-}$(Oxidation)

Overall reaction $2 PbSO_4(s)+2 H_2 O \rightarrow Pb(s)+PbO_2(s)+4 H^{+}$(aq. $)+2 SO_4{ }^{2-}$ (aq.)

• Option (c) is incorrect because during the charging process, the $PbSO_4$ at the cathode is reduced to $Pb$, not oxidized. Oxidation would involve the loss of electrons, but in this case, the cathode gains electrons.

• Option (b) is incorrect because it states the correct process for the cathode but is listed as option (b) instead of option (a). The correct reduction reaction at the cathode is indeed $PbSO_4$ being reduced to $Pb$.

• Option (d) is incorrect because during the charging process, the $PbSO_4$ at the anode is oxidized to $PbO_2$, not $Pb$. The anode reaction involves the conversion of $PbSO_4$ to $PbO_2$, which is an oxidation process.

Thinking Process

This question is based on the concept of Kohlrausch law and can be solved by using the concept involved in calculation of limiting molar conductivity of any salt. According to Kohlrausch law limiting molar conductivity of any salt is equal to sum of limiting molar conductivity of individual molar conductivity of cations and anions of electrolyte.

Answer

(b)

$$ \begin{array}{llllll} & \Lambda_{m(NH_4 Cl)}^{\circ} &= &\Lambda_{m(NH_4^{+})}^{\circ} &+ &\Lambda_{m(Cl^{-})}^{\circ} \\ & \Lambda_{m(NaOH)}^{\circ} &=& \Lambda_{m(Na^{+})}^{\circ}& + &\Lambda_{m(OH^{-})}^{\circ} \\ & \Lambda_{m(NaCl)}^{\circ}& = &\Lambda_{m(Na^{+})}^{\circ} &+ &\Lambda_{m(Cl^{-})}^{-} \\ & - & &- & &- \\ \hline & \Lambda_{m(NH_4 Cl)}^{\circ} & + &\Lambda_{m(NaOH)}^{\circ}-\Lambda_{m(NaCl)}^{\circ}& = &\Lambda_{m(NH_4 OH)}^{\circ} \end{array} $$

Hence, option (b) is correct choice.

• Option (a): This option is incorrect because it incorrectly includes $\Lambda_{m(NH_4 OH)}^{\circ}$ on both sides of the equation, which is not mathematically valid. The correct approach should involve combining the molar conductivities of different ions to form $\Lambda_{m(NH_4 OH)}^{\circ}$, not including it directly.

• Option (c): This option is incorrect because it incorrectly combines the molar conductivities of $\Lambda_{m(NH_4 Cl)}^{\circ}$, $\Lambda_{m(NaCl)}^{\circ}$, and subtracts $\Lambda_{m(NaOH)}^{\circ}$. This combination does not correctly account for the ions present in $NH_4OH$. Specifically, it does not correctly balance the contributions of $NH_4^+$ and $OH^-$ ions.

• Option (d): This option is incorrect because it incorrectly combines the molar conductivities of $\Lambda_{m(NaOH)}^{\circ}$, $\Lambda_{m(NaCl)}^{\circ}$, and subtracts $\Lambda_{m(NH_4 Cl)}^{\circ}$. This combination does not correctly account for the ions present in $NH_4OH$. Specifically, it does not correctly balance the contributions of $NH_4^+$ and $OH^-$ ions.

Answer

(d) In case of electrolysis of aqueous $NaCl$ oxidation reaction occurs at anode as follows

$$ \begin{matrix} Cl^{-}(aq) \longrightarrow \frac{1}{2} Cl_2(g)+e^{-} & E^{\circ}=1.36 V \\ 2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} . & E_cell^{\circ}=1.23 V \end{matrix} $$

But due to lower $E_{\text {cell }}^{\circ}$ value water should get oxidised in preference of $Cl^{-}(a q)$.

However, the actual reaction taking place in the concentrated solution of $NaCl$ is (d) and not (b) i.e., $Cl_2$ is produced and not $O_2$.

This unexpected result is explained on the basis of the concept of ‘overvoltage’, i.e., water needs greater voltage for oxidation to $O_2$ (as it is kinetically slow process) than that needed for oxidation of $Cl^{-}$ions to $Cl_2$. Thus, the correct option is (d) not (b).

• (a) $Na^{+}(aq)+e^{-} \longrightarrow Na(s) ; E_{\text {cell }}^{s}=-2.71 V$: This reaction represents the reduction of sodium ions to sodium metal, which would occur at the cathode, not the anode. The anode is where oxidation occurs.

• (b) $2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; E_{\text {cell }}^{s}=-1.23 V$: Although this is an oxidation reaction that could occur at the anode, in the case of concentrated NaCl solution, the oxidation of chloride ions to chlorine gas is preferred due to the overvoltage effect. Water oxidation is kinetically slower and requires a higher voltage.

• (c) $H^{+}(aq)+e^{-} \longrightarrow \frac{1}{2} H_2(g)$; $E_{\text {cell }}^{-}=0.00 V$: This reaction represents the reduction of hydrogen ions to hydrogen gas, which would occur at the cathode, not the anode. The anode is where oxidation occurs.

Answer

$(b, d)$

‘Lesser the $E^{\circ}$ value of redox couple higher the reducing power"

$$ \begin{matrix} Cu^{2+}+2 e^{-} \longrightarrow Cu & E^{\circ}=0.34 V \\ 2 H^{+}+2 e^{-} \longrightarrow H_2 & E^{\circ}=0.00 V \end{matrix} $$

Since, $2 H^{+} / H_2$ has lesser SRP than $Cu^{2+} / Cu$ redox couple. Therefore,

(i) This redox couple is a stronger oxidising agent than $H^{+} / H_2$.

(ii) $Cu$ can’t displace $H_2$ from acid.

Hence, (b) and (d) are correct.

• (a) this redox couple is a stronger reducing agent than the $H^{+} / H_2$ couple: This is incorrect because a higher standard electrode potential ($E^{\circ}$) indicates a stronger oxidizing agent, not a stronger reducing agent. Since $Cu^{2+} / Cu$ has a higher $E^{\circ}$ value than $H^{+} / H_2$, it means $Cu^{2+} / Cu$ is a stronger oxidizing agent, not a stronger reducing agent.

• (c) $Cu$ can displace $H_2$ from acid: This is incorrect because for $Cu$ to displace $H_2$ from acid, $Cu$ would need to be a stronger reducing agent than $H^{+} / H_2$. However, since $Cu^{2+} / Cu$ has a higher $E^{\circ}$ value, it is a weaker reducing agent compared to $H^{+} / H_2$. Therefore, $Cu$ cannot displace $H_2$ from acid.

Answer

(a, c)

During the electrolysis of dilute sulphuric acid above given three reaction occurs each of which represents particular reaction either oxidation half cell reaction or reduction half cell reaction.

Oxidation half cell reactions occur at anode are as follows

$$ \begin{aligned} 2 SO_4^{2-}(aq) \longrightarrow S_2 O_8^{2-}+2 e^{-} & E_{\text {cell }}^{\circ}=1.96 V \\ 2 H_2 O^{+}(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; & E_{\text {cell }}^{\circ}=1.23 V \end{aligned} $$

Reaction having lower value of $E_cell^{\circ}$ will undergo faster oxidation.

Hence, oxidation of water occur preferentially reduction half cell reaction occurs at cathode

$$ H^{+}(aq)+e^{-} \longrightarrow \underset{\text{At cathode}}{\frac{1}{2} H_2(g)} $$

Hence, options (a) and (b) are correct.

• Option (b) is incorrect: In concentrated sulfuric acid solution, the oxidation of sulfate ions ($SO_4^{2-}$) to peroxodisulfate ($S_2O_8^{2-}$) is more favorable due to its higher standard electrode potential ($E_{\text{cell}}^{\circ} = 1.96 , \text{V}$) compared to the oxidation of water ($E_{\text{cell}}^{\circ} = 1.23 , \text{V}$). Therefore, sulfate ions will be oxidized at the anode rather than water.

• Option (d) is incorrect: In dilute sulfuric acid solution, the oxidation of water to oxygen gas ($O_2$) is more favorable due to its lower standard electrode potential ($E_{\text{cell}}^{\circ} = 1.23 , \text{V}$) compared to the oxidation of sulfate ions to peroxodisulfate ($E_{\text{cell}}^{\circ} = 1.96 , \text{V}$). Therefore, water will be oxidized at the anode rather than sulfate ions being oxidized to tetrathionate ions.

Answer

$(b, c)$

At state of equilibrium

$$ \begin{aligned} \Delta G & =-R T \log K \\ -n F E^{\circ} & =-R T 2.303 \log K_{C} \\ E^{\circ} & =\frac{+R T 2.303 \log K_{C}}{+2 F} \quad(n=2 \text { for Daniel cell }) \end{aligned} $$

$\because$ At equilibrium $E^{\circ}=1.1$

$$ \begin{aligned} & \therefore \quad \frac{2.303 R T}{2 F} \log K_{C}=1.1 \\ & \log K_{C}=\frac{2.2}{0.059} \quad \text{[on solving]} \end{aligned} $$

Hence, options (b) and (c) are the correct choices.

• Option (a) is incorrect because (1.1) is the standard cell potential ((E_{\text{cell}}^{\circ})) and not the equilibrium constant ((K_C)). The equilibrium constant is a dimensionless quantity and cannot be directly equated to a voltage.

• Option (d) is incorrect because (\log K_C) is not equal to the standard cell potential ((E_{\text{cell}}^{\circ})). The correct relationship involves the logarithm of the equilibrium constant being proportional to the standard cell potential, as shown in the correct expressions (b) and (c).

Answer

$(a, b)$

Conductivity of electrolytic solution is due to presence of mobile ions in the solution. This type of conductance is known as ionic conductance. Conductivity of these type of solutions depend upon

(i) the nature of electrolyte added

(ii) size of the ion produced and their solvatian

(iii) concentration of electrolyte

(iv) nature of solvent and its viscosity

(v) temperature

While power of source or distance between electrodes has no effect on conductivity of electrolyte solution.

Hence, options (a) and (b)are the correct choices.

• Power of AC source: The power of the AC source does not affect the conductivity of the electrolytic solution because conductivity is determined by the intrinsic properties of the solution, such as the nature and concentration of the electrolyte, and not by the external power applied.

• Distance between the electrodes: The distance between the electrodes does not affect the conductivity of the electrolytic solution because conductivity is a property of the solution itself. While the distance between electrodes can affect the resistance measured in a specific setup, it does not change the inherent conductivity of the solution.

Thinking Process

This problem includes concept of Kohlrausch law and its application in determination of molar conductivity of species. This problem can be solved by following three steps.

(i) Write the molar conductance of each species in terms of sum of their constituent ions.

(ii) Now operate the equation of each option given above by using information provided in the question.

(iii) At last if the sum of molar conductivity remaining constituent ions is equal to the molar conductivity of species asked (here $\Lambda_{m(H_2 O)}^{0}$ ) then that will be the correct choice.

Answer

(a, c)

(a)

$$ \begin{array}{lllllll} \Lambda _{m(HCl)}^{\circ} & =& \Lambda _{m(H^{+})}^{\circ}& +& \Lambda _{m(Cl^{-})}^{\circ} &\\ \Lambda _{m(HCl)}^{\circ} & =& \Lambda _{m(H^{+})}^{\circ}& +& \Lambda _{m(Cl^{-})}^{\circ} &\\ \Lambda _{m(NaCl)}^{\circ} & =& \Lambda _{m(Na^{+}) }^{0}& +& \Lambda _{m(Cl^{-})} & \\ && - &&- \\ \hline \Lambda _{m\left(\mathrm{HCl}\right)}^{\circ} &+& \Lambda _{m(\mathrm{NaOH})}^{\circ}-\Lambda _{m(\mathrm{NaCl})}^{\circ} & = & \Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ} +\Lambda _{m\left(\mathrm{HO}^{-}\right)}^{\circ}\\ & &&& =\Lambda _{m\left(\mathrm{H}_2 \mathrm{O}\right)} \end{array} $$

(d)

$$ \begin{array}{llll} \Lambda _{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ} & = &\Lambda _{m\left(\mathrm{NH}_4{ }^{+}\right)}^{\circ} & +& \Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ} \\ \Lambda _{m(\mathrm{HCl})}^{\circ} & =& \Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ}& +& \Lambda _{m\left(\mathrm{Cl}^{-}\right)}^{\circ} \\ \Lambda _{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}& =& \Lambda _{m\left(\mathrm{NH}_4{ }^{+}\right)}^{\circ}& +& \Lambda _{m\left(\mathrm{Cl}^{-}\right)}^{\circ} \\ \text{-} & & - && - \\ \hline \Lambda _{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}& +& \Lambda _{m(\mathrm{HCl})}^{\circ} &-& \Lambda _{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ} \\ & & =\Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ} &+&\Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ}=\Lambda _{m\left(\mathrm{H}_2 \mathrm{O}\right)}^{\circ} \end{array} $$

This type of decomposition is not possible due to weak basic strength of $NH_4 OH$. This line will be placed above.

(b)

$$ \begin{array}{ll} \Lambda _{m\left(\mathrm{HNO}_3\right)}^{\circ} & =\Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{NO}_3^{-}\right)}^{\circ} \\ \Lambda _{m\left(\mathrm{NaNO}_3\right)}^{\circ} & =\Lambda _{m\left(\mathrm{Na}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{NO}_3^{-}\right)}^{\circ} \\ \Lambda _{m(\mathrm{NaOH})}^{\circ} & =\Lambda _{m\left(\mathrm{Na}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ} \\ \text{-} & - \\ \hline \Lambda _{m\left(\mathrm{HNO}_3\right)}^{\circ} & +\Lambda _{m\left(\mathrm{NaNO}_3\right)}^{\circ}-\Lambda _{m(\mathrm{NaOH})}^{\circ} \\ & =\Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ}+2 \Lambda _{m\left(\mathrm{NO}_3^{-}\right)}^{\circ}-\Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ} \end{array} $$

(b) is incorrect

(c)

$$ \begin{array}{ll} \Lambda _{m\left(\mathrm{HNO} _3\right)}^{\circ} & =\Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{NO} _3^{-}\right)}^{\circ} \\ \Lambda _{m\left(\mathrm{NaOH}^{\circ}\right)}^{\circ} & =\Lambda _{m\left(\mathrm{Na}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ} \\ \Lambda _{m\left(\mathrm{NaNO} _3\right)}^{\circ} & =\Lambda _{m\left(\mathrm{Na} _{-}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{NO} _3^{-}\right)}^{\circ} \\ \mathrm{-} & \mathrm{-} \\ \hline \Lambda _{m\left(\mathrm{HNO} _3\right)}^{\circ} & +\Lambda _{m(\mathrm{NaOH})}^{\circ}-\Lambda _{m\left(\mathrm{NaNO} _3\right)}^{\circ} \\ & =\Lambda _{m\left(\mathrm{H}^{+}\right)}^{\circ}+\Lambda _{m\left(\mathrm{OH}^{-}\right)}^{\circ}=\Lambda _{m\left(\mathrm{H} _2 \mathrm{O}\right)}^{\circ} \end{array} $$

Hence, Options (a) and (c) are the correct choices.

• Option (b) is incorrect because the expression (\Lambda_{m(HNO_3)}^{\circ} + \Lambda_{m(NaNO_3)}^{\circ} - \Lambda_{m(NaOH)}^{\circ}) simplifies to (\Lambda_{m(H^{+})}^{\circ} + 2\Lambda_{m(NO_3^{-})}^{\circ} - \Lambda_{m(OH^{-})}^{\circ}), which does not result in (\Lambda_{m(H_2O)}^{\circ}). Instead, it includes an extra term (\Lambda_{m(NO_3^{-})}^{\circ}).

• Option (d) is incorrect because the decomposition involving (\Lambda_{m(NH_4OH)}^{\circ}) is not valid due to the weak basic strength of (NH_4OH). The expression (\Lambda_{m(NH_4OH)}^{\circ} + \Lambda_{m(HCl)}^{\circ} - \Lambda_{m(NH_4Cl)}^{\circ}) would theoretically simplify to (\Lambda_{m(H^{+})}^{\circ} + \Lambda_{m(OH^{-})}^{\circ}), but the weak basic nature of (NH_4OH) makes this decomposition impractical.

Thinking Process

This problem is based on electrolysis of electrolytes.

Answer

(a, c)

For electrolysis of aqueous solution of $CuSO_4$.

$$ \begin{aligned} CuSO_4(aq) & \longrightarrow Cu^{2+}+SO_4^{2-} \\ H_2 O & \longrightarrow 2 H^{+}+O^{2-} \end{aligned} $$

$$ \text{At anode} \quad 2 O^{2-} \longrightarrow O_2+2 e^{-} $$

$$ \text{At cathode} \quad Cu^{2+}+2 e^{-} \longrightarrow Cu(s) $$

• (b) Copper will deposit at anode: This is incorrect because, in the electrolysis of an aqueous solution of $CuSO_4$ using platinum electrodes, the anode reaction involves the oxidation of water to produce oxygen gas, not the deposition of copper. Copper deposition occurs at the cathode.

• (d) Copper will dissolve at anode: This is incorrect because the anode reaction in this setup involves the oxidation of water to release oxygen gas. Copper dissolution at the anode would occur if the anode itself were made of copper, but in this case, platinum electrodes are used.

Answer

$(a, b)$

Electrolysis of $CuSO_4$ can be represented by two half cell reactions these occurring at cathode and anode respectively as

$\text{At anode} \quad Cu^{2+}+2 e^{-} \longrightarrow Cu(s)$

$ \text{At cathode} \quad Cu(s) \longrightarrow Cu^{2+}+2 e^{-}$

Here, Cu will deposit at cathode while copper will dissolved at anode.

Hence, options (a) and (b) are the correct choices.

• Option (c): Oxygen will be released at anode

  • This option is incorrect because in the presence of copper electrodes, the copper anode will dissolve into the solution as Cu²⁺ ions rather than releasing oxygen. The dissolution of copper at the anode is more favorable than the oxidation of water to release oxygen.

• Option (d): Copper will deposit at anode

  • This option is incorrect because at the anode, copper actually dissolves into the solution as Cu²⁺ ions. Deposition of copper occurs at the cathode, not the anode.

Answer

$(a, b)$

As we know that, conductance is reciprocal of resistance and conductivity is the conductance of $1 cm^{3}$ of substance. Also,conductivity is reciprocal of resistivity.

$$\kappa =\frac{1}{\rho} $$

$$ R =\rho \frac{l}{A} $$

$$ \rho =\frac{R \cdot A}{l} \Rightarrow \kappa=\frac{1}{(\frac{R \cdot A}{l})} $$

$$ \kappa \frac{1}{R} \cdot \frac{l}{A} = \frac{1}{R} \times G^* = \frac{G^*}{R} $$

Hence, options (a) and (b) are the correct choices.

• Option (c) $\Lambda_{m}$ is incorrect because $\Lambda_{m}$ represents molar conductivity, which is different from conductivity $\kappa$. Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, whereas conductivity $\kappa$ is a measure of a material’s ability to conduct electric current.

• Option (d) $\frac{l}{A}$ is incorrect because it represents the ratio of the length to the cross-sectional area of a material, which is related to the geometry of the material rather than its intrinsic property of conductivity. Conductivity $\kappa$ is a material property and is not directly given by the ratio $\frac{l}{A}$.

Answer

(a, c)

Molar conductivity is the conductivity due to ions furnished by one mole of electrolyte in solution. Molar conductivity of ionic solution depends on

(i) Temperature Molar conductivity of electrolyte solution increases with increase in temperature.

(iii) Concentration of electrolytes in solution As concentration of electrolyte increases, molar conductivity decreases.

$$ \therefore \quad \lambda=\frac{\kappa}{c} $$

• Distance between electrodes: The distance between electrodes affects the overall resistance of the solution but does not directly influence the molar conductivity, which is an intrinsic property of the electrolyte solution.

• Surface area of electrodes: The surface area of electrodes can affect the current and the measured conductivity in practical setups, but it does not change the intrinsic molar conductivity of the solution itself.

Answer

$(b, c)$

Left side of cell reaction represents oxidation half cell i.e., oxidation of $Mg$ and right side of cell represents reduction half cell reactions i.e., reduction of copper.

(ii) $Cu$ is reduced and reduction occurs at cathode.

(iii) $Mg$ is oxidised and oxidation occurs at anode.

(iv) whole cell reaction can be written as

Hence, options (b) and (c) both are correct choices.

• Option (a) is incorrect because in the given cell, $Mg$ is the anode where oxidation occurs, not the cathode. The cathode is where reduction occurs, and in this case, it is $Cu$.

• Option (d) is incorrect because $Cu$ is the species being reduced, which means it is the oxidizing agent. However, the statement is misleading as it does not clearly distinguish between the element $Cu$ and the ion $Cu^{2+}$. The correct statement should be that $Cu^{2+}$ is the oxidizing agent.

Answer

No, only the difference in potential between two electrodes can be measured. This is due to the reason that oxidation or reduction cannot occur alone. So, when we measure electrode potential of any electrode we have to take a reference electrode.

Answer

No, otherwise the reaction become non-feasible.

The reaction is feasible only at $E_{\text {cell }}^{\circ}=$ positive or $\Delta_{r} G^{\circ}=$ negative .

when $E^{\circ}=\Delta_{r} G^{\circ}=0$ the reaction reaches at equilibrium.

Answer

At the stage of chemical equilibrium in the cell.

$$ \begin{aligned} E_{cell} & =0 \\ \Delta_{r} G^{\circ} & =-n F E_cell^{\circ} \\ & =-n \times F \times 0=0 \end{aligned} $$

Answer

Greater the negative reactivity of standard reduction potential of metal greater is its reactivity. It means that $Zn$ is more reactive than hydrogen. When zinc electrode will be connected to standard hydrogen electrode, $Zn$ will get oxidised and $H^{+}$will get reduced.

Thus, zinc electrode will be the anode of the cell and hydrogen electrode will be the cathode of the cell.

Answer

Different masses of $Cu$ and $Ag$ will be deposited at cathode. According to Faraday’s second law of electrolysis amount of different substances liberated by same quantity of electricity passes through electrolyte solution is directly proportional to their chemical equivalent weight.

$$ \frac{W_1}{W_2}=\frac{E_1}{E_2} $$

where, $E_1$ and $E_2$ have different values depending upon number of electrons required to reduce the metal ion. Thus, masses of $Cu$ and $Ag$ deposited will be different.

Answer

In a galvanic cell, oxidation half reaction is written on left hand side and reduction half reaction is on right hand side. Salt bridge is represented by parallel lines $Cu|Cu^{2+} || Ag^{+}| Ag$.

Answer

Under the condition of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential. $So, Cl^{-}$is oxidized at anode instead of water.

Possible oxidation half cell reactions occurring at anode are

$$ \begin{matrix} Cl^{-}(aq) \longrightarrow \frac{1}{2} Cl_2(g)+e^{-} ; & E_cell^{s}=1.36 V \\ 2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; & E_cell^{\circ}=1.23 V \end{matrix} $$

Species having lower $E_{\text {cell }}^{\circ}$ cell undergo oxidation first than the higher value but oxidation of $H_2 O$ to $O_2$ is kinetically so slow that it needs some overvoltage.

Answer

The potential difference between the electrode and the electrolyte in an electrochemical cell is called electrode potential.

Answer

The cell drawn above represents electrochemical cell in which two different electrodes are fitted in their respective electrolytic solution and cell drawn at bottom represents electrolytic cell.

Cell representation can be represented as $Zn|Zn^{2+} || Cu^{2+}| Cu$.

$Zn$ is loosing electrons which are going towards electrode (A) and copper is accepting electron from electrode $B$. Hence,

Charge on electrode $A=+$

Charge on electrode $B=-$

Answer

Alternating current is used in electrolysis so that concentration of ions in the solution remains constant and exact value of resistance is measured.

Answer

When an opposing potential of $1.1 V$ is applied to a galvanic cell having electrical potential of $1.1 V$ then cell reaction stops completely and no current will flow through the cell.

Answer

The $pH$ of the solution will rise as $NaOH$ is formed in the electrolytic cell.

Chemical reaction occurring at cell when aqueous brine solution is electrolysed are as follows

$$ \begin{aligned} & NaCl(aq) \stackrel{H_2 O}{\longrightarrow} Na^{+}(aq)+Cl^{-}(aq) \\ & \text { Cathode } H_2 O(l)+e^{-} \longrightarrow \frac{1}{2} H_2(g)+OH^{-}(aq) \\ & \text { Anode } Cl^{-}(aq) \longrightarrow \frac{1}{2} Cl_2(g)+e^{-} \\ & \text{Net reaction} \quad NaCl(aq)+H_2 O(l) \longrightarrow Na^{+}(aq)+OH^{-}(aq)+\frac{1}{2} H_2+\frac{1}{2} Cl_2 \end{aligned} $$

Answer

Life time of any cell depends upon ions present in cell. lons are not involved in the overall cell reaction of mercury cell. Hence, mercury cell has a constant cell potential throughout its useful life.

Answer

Strong electrolytes dissociate almost completely even on high concentration. Therefore, concentration of such solutions remain almost same on dilution. Electrolyte ’ $B$ ’ is stronger than ’ $A$ ’ because in ’ $B$ ’ the number of ions remains the same on dilution, but only interionic attraction decreases.

Therefore, $\Lambda_{m}$ increases only 1.5 times. While in case of weak electrolyte on dilution, number of constituent ions increases.

Answer

Since $pH$ of solution depends upon concentration of $H^{+}$presence in solutions. $pH$ of the solution will not be affected as $[H^{+}]$remains constant.

$$ \begin{gathered} \text { At anode } 2 H_2 O \longrightarrow O_2+4 H^{+}+4 e^{-} \\ \text {At cathode } 4 H^{+}+4 e^{-} \longrightarrow 2 H_2 \end{gathered} $$

Answer

Conductivity of solution due to total ions present in per unit volume of solution is known as specific conductivity. Specific conductivity decreases due to decrease in the number of ions per unit volume. We add water to aqueous solution, number of ions present in per unit volume decreases.

Answer

Standard hydrogen electrode (SHE) is the reference electrode whose electrode potential is taken to be zero. The electrode potential of other electrodes are measured with respect to it .

Answer

Cell reaction represented in the question is composed of two half cell reactions. These reactions are as follows

$$ \begin{aligned} & \text { At anode } Cu \longrightarrow Cu^{2+}+2 e^{-} \\ & \text {At cathode } Cl_2+2 e^{-} \longrightarrow 2 Cl^{-} \end{aligned} $$

Copper is getting oxidised at anode. $Cl_2$ is getting reduced at cathode.

Answer

$Zn+Cu^{2+} \longrightarrow Zn^{2+}+Cu$

$$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log [\frac{Zn^{2+}}{Cu^{2+}}] \\ & E_{\text {cell }}^{\circ}=E_{\text {cell }}^{\circ}+\frac{0.0591}{2} \log [\frac{Cu^{2+}}{Zn^{2+}}] \end{aligned} $$

According to this equation

$E_{\text {cell }}^{\circ}$ is directly dependent on concentration of $Cu^{2+}$ and inversely dependent upon concentration of $Zn^{2+}$ ions.

$E_{\text {cell }}$ decreases when concentration of $Zn^{2+}$ ions is increased.

Answer

Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but it take a long time. Fuel cell run continuously as long as the reactants are supplied to it and products are removed continuously.

Answer

When a lead storage battery is discharged then the following cell reaction takes place

$$ Pb+PbO_2+2 H_2 SO_4 \longrightarrow 2 PbSO_4+2 H_2 O $$

Density of electrolyte depends upon number. of constituent ions present in per unit volume of electrolyte solution. In this case density of electrolyte decreases as water is formed and sulphuric acid is consumed as the product during discharge of the battery.

Answer

In the case of $CH_3 COOH$, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.

$$ CH_3 COOH+H_2 O \rightleftharpoons CH_3 COO^{-}+H_3 O^{+} $$

In case of strong electrolyte, the number of ions remains the same but the interionic attraction decreases.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

D. $\rightarrow(3)$

A. $\wedge_{m}$ (molar conductivity) is the conductivity due to number of ions furnished by one mole of electrolyte. As dilution increases number of ions present in the solution increases hence molar conductivity increases.

B. $E_{\text {cell }}^{\circ}$ of any atom/ion does not depend upon number of atom/ion, hence $E_{\text {cell }}^{\circ}$ of any atom/ion is an intensive properties.

C. $\kappa$ represents specific conductivity which depends upon number of ions present in per unit volume.

D. $\Delta_{r} G_{\text {cell }}$ is an extensive property as it depends upon number of particles(species).

Answer

A. $\rightarrow(4)$

B. $\rightarrow$ (3)

C. $\rightarrow(1)$

D. $\rightarrow$ (2)

A. Chemical reaction occurring on lead storage battery can be represented as

At anode $\quad Pb(s)+SO_4{ }^{2-}(aq) \longrightarrow PbSO_4(s)+2 e^{-}$

At cathode $ PbO_2(s)+SO_4^{2-}(aq)+4 H^{+}(aq) \xrightarrow{+2 e^-} 2 PbSO_4(s)+2 H_2 O(l) $

Thus, $Pb$ is anode and $PbO_2$ is cathode.

B. Mercury cell does not include ions during their function hence produce steady current.

C. Fuel cell has maximum efficiency as they produce energy due to combustion reaction of fuel.

D. Rusting is prevented by corrosion.

Answer

A. $\rightarrow(4)$

B. $\rightarrow$ (3)

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Conductivity $(\kappa)=\frac{G^{*}}{R}$

B. Molar conductivity $(\wedge_{m})=\frac{\kappa}{C}$

C. Degree of dissociation $(\alpha)=\frac{\wedge_{m}}{\wedge_m^{\circ}}$

D. Charge $Q=I \times t$

where, $Q$ is the quantity of charge in coulomb when $I$ ampere of current is passed through an electrolyte for $t$ second.

Answer

A. $\rightarrow(4)$

B). $\rightarrow$ (3)

C. $\rightarrow(1,5)$

D. $\rightarrow(2)$

A. Lechlanche cell The electrode reaction occurs at Lechlanche cell are

$$ \text { At anode } Zn(s) \longrightarrow Zn^{2}+2 e^{-} $$

At cathode $MnO_2+NH_4^{+}+e^{-} \longrightarrow MnO(OH)+NH_3$

B. Ni-Cd cell is rechargeable. So, it has more life time.

C. Fuel cell produces energy due to combustion. So, fuel cell converts energy of combustion into electrical energy e.g., $2 H_2+O_2 \longrightarrow 2 H_2 O$

D. Mercury cell does not involve any ion in solution and is used in hearing aids.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(7)$

D. $\rightarrow(5)$

E. $\rightarrow(4)$

F. $\rightarrow(2)$

G. $\rightarrow(6)$

A. $F_2$ is a non-metal and best oxidising agent because SRP of $F_2$ is $+2.87 V$.

B. $Li$ is a metal and strongest reducing agent because SRP of $Li$ is $-3.05 V$.

C. $Au^{3+}$ is a metal ion which is an oxidising agent as SRP of $Au^{3+}$ is $+1.40 V$.

D. $Br^{-}$is an anion that can be oxidised by

$Au^{3+}$ as $Au^{3+}(E^{\circ}=1.40)$ is greater than

$$ Br^{-}(E^{\circ}=1.09 V) $$

E. Au is an unreactive metal.

F. $Li^{+}$is a metal ion having least value of $SRP(-3.05 V)$, hence it is the weakest oxidising agent.

G. $F^{-}$is an anion which is the weakest reducing agent as $F^{-} / F_2$ has low oxidation potential $(-2.87 V)$.

Answer

(c) Assertion is true but the reason is false. Electrode potential of $Cu^{2+} / Cu$ is $+0.34 V$ and Electrode potential of $2 H^{+} / H_2$ is $0.00 V$.

Hence, correct reason is due to positive value of $Cu^{2+} / Cu$ it looses electron to $H^{+}$and get reduces, while $H_2$ gas evolves out.

Answer

(c) Assertion is true but the reason is false. Feasibility of chemical reaction depends on Gibbs free energy which is related to $E_{\text {cell }}^{\circ}$ as

$$ \Delta G^{-}=-nFE-_{\text {cell }} $$

When value of $E_{\text {cell }}^{s}$ is positive then $\Delta G^{s}$ becomes negative. Hence, reaction becomes feasible.

Correct reason is $E_{\text {cathode }}>E_{\text {anode }}$.

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion. Since, conductivity depends upon number of ions per unit volume. Therefore, the conductivity of all electrolytes decreases on dilution due to decrease in number of ions per unit volume.

Answer

(a) Assertion and reason are true and the reason is the correct explanation of the assertion.

Molar conductivity of weak electrolytic solution increases on dilution, because as we add excess water to increase the dilution degree of dissociation increases which lead to increase in number of ions in the solution. Thus, $\Lambda_{m}$ show a very sharp increase.

Answer

(e) Assertion is false but reason is true. Correct assertion is mercury cell gives steady potential.

Reason is correct as ions are not involved in cell reaction.

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Explanation Electrolysis of $NaCl$ is represented by following chemical reactions At cathode

$$ H^{+}(aq)+e^{-} \longrightarrow \frac{1}{2} H_2(g) $$

$$ \begin{aligned} \text{At anode} \quad Cl^{-}(aq) & \longrightarrow \frac{1}{2} Cl_2+e^{-} ; E_cell^{\circ}=1.36 V \\ 2 H_2 O(aq) & \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; E_Cell^{\circ}=1.23 V \end{aligned} $$

$E_{\text {cell }}^{\circ}$ for this reaction has lower value but formation of oxygen at anode requires over potential.

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion

Concentration of ionic solution changes on using DC current as a source of energy while on passing AC current concentration does not change. Hence, AC source is used for measuring resistance

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Current stop flowing when $E_{\text {cell }}=0$

As at $E_{cell}=0$ reaction reaches the equilibrium.

Answer

(b) Both assertion and reason are correct but reason is not the correct explanation of assertion.

$$ \begin{aligned} & E=E-\frac{0.0591}{1} \log \frac{1}{[Ag^{+}]} \\ & E=E^{\circ}+0.059 \log [Ag^{+}] \end{aligned} $$

Thus, $E_{Ag^{+} / Ag}$ increases with increase in concentration of $Ag^{+}$.

Answer

(d) Both assertion and reason are false.

Copper sulphate can’t be stored in zinc vessel as zinc is more reactive than copper due to negative value of standard reduction potential of $Zn$.

Thinking Process

This problem includes concept of electrochemical cell, electrolytic cell and charge on electrode. To solve this problem identify the charge on each electrode first.

Answer

(i) Cell ’ $B$ ’ will act as electrolytic cell due to its lesser value of emf.

The electrode reactions will be

At cathode

$$ Zn^{2+}+2 e^{-} \longrightarrow Zn $$

At anode

$$ Cu \longrightarrow Cu^{2+}+2 e^{-} $$

(ii) If cell ’ $B$ ’ has higher emf, it acts as galvanic cell.

Now it will push electrons into cell ’ $A$ '

In this case, the reactions will be

$$ \begin{aligned} Zn & \longrightarrow Zn^{2+}+2 e^{-} \text {(At anode) } \\ Cu^{2+}+2 e^{-} & \longrightarrow Cu \text { (At cathode) } \end{aligned} $$

Answer

(i) Electrons move from $Zn$ to $Ag$ as $E^{\circ}$ is more negative for $Zn$, so $Zn$ undergoes oxidation and $Ag^{+}$undergoes reduction.

(ii) $Ag$ is the cathode as it is the site of reduction where $Ag^{+}$takes electrons from medium and deposit at cathode.

(iii) Cell will stop functioning because cell potential drops to zero. At $E=0$ reaction reaches equilibrium. (iv) When $E_{\text {cell }}=0$ because at this condition reaction reaches to equilibrium.

(v) Concentration of $Zn^{2+}$ ions will increase and concentration of $Ag^{+}$ions will decrease because $Zn$ is converted into $Zn^{2+}$ and $Ag^{+}$is converted into $Ag$.

(vi) When $E_{\text {cell }}=0$ equilibrium is reached and concentration of $Zn^{2+}$ ions and $Ag^{+}$will not change.

Answer

If concentration of all reacting species is unity, then $E _{\text {cell }}=E _{\text {cell }}^{\circ}$ and $\Delta G^{\circ}=-n F E^{\circ} _{\text {cell }}$ where, $\Delta _{r} G^{\circ}$ is standard Gibbs energy of the reaction

$$ \begin{aligned} E_{\text {cell }}^{\circ} & =\text { emf of the cell } \\ n F & =\text { charge passed } \end{aligned} $$

If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.

The reversibly work done by a galvanic cell is equal to decrease in its Gibbs energy.

$$ \Delta_{r} G=-n F E_{\text {cell }} $$

As $E_{\text {cell }}$ is an intensive parameter but $\Delta_{r} G$ is an extensive thermodynamic property and the value depends on $n$.

For reaction, $Zn(s)+Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq)+Cu(s)$ in a galvanic cell.

$$\Delta_{r} G=-2 F E_{\text {cell }}$$ [Here, $n=2$ ]