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.

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

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

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).

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.)

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.

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).

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.

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.

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.

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.

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.

Thinking Process

This problem is based on electrolysis of electrolytes.

Answer

(a, c)

For electrolysis of aqueous solution of $CuSO_4$.

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

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

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

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.

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.

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.

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

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.

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, $n=2$ ]