Chemistry-class-12-html_c-12-electrochemistry

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**1.** Which cell will measure standard electrode potential of copper electrode?

(a) Pt (s) $\mid H_2(g, 0.1.$ bar) $\mid H^{+}($aq., $1 M)|| Cu^{2+}(aq, 1 M) \mid Cu$

(b) Pt (s) $\mid H_2(g, 1.$ bar) $|H^{+}(aq, 1 M)||Cu^{2+}(aq, 2 M)| Cu$

(d) Pt(s) $\mid H_2(g, 0.1$ bar $)|H^{+}(aq, 0.1 M) \| Cu^{2+}(aq, 1 M)| Cu$

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**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 }} $

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**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

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$
\begin{matrix} 
E   _{Mg^{2+} / Mg} & =(\frac{0.059}{2}) & \log [Mg^{2+}] & +E _{Mg^{2+} / Mg}^{\circ} \\\\\\
\uparrow & \uparrow & \uparrow & \uparrow \\\\\\
\text{ Y }& \text{ M } & \text{ X } & \text{ +C }
\end{matrix} 
$

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

Thus, equation can be diagramatically represented as

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**3.** Which of the following statement is correct?

(a) $E   _{\text{cell }}$ and $\Delta_r G$ of cell reaction both are extensive properties

(b) $E   _{\text{cell }}$ and $\Delta_r G$ of cell reaction both are intensive properties

(d) $E   _{\text{cell }}$ is an extensive property while $\Delta_r G$ of cell reaction is an intensive property

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**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.

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**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.

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**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** $PbSO_4(s)+2 e^{-} \to Pb(s)+SO_4{ }^{2-}(aq)$ (Reduction)

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

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

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**16.** $ \Lambda   _{m(NH_4 OH)}^{\circ}$ is equal to...... .

(a) $\Lambda   _{m(NH_4 OH)}^{\circ}+\Lambda _{m(NH_4 Cl)}^{\circ}-\Lambda _{(HCl)}^{\circ}$

(b) $\Lambda   _{m(NH_4 Cl)}^{\circ}+\Lambda _{m(NaOH)}^{\circ}-\Lambda _{(NaCl)}^{\circ}$

(d) $\Lambda   _{m(NaOH)}^{\circ}+\Lambda _{m(NaCl)}^{\circ}-\Lambda _{(NH_4 Cl)}^{\circ}$

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**17.** In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

(a) $Na^{+}(aq)+e^{-} \longrightarrow Na(s) ; E   _{\text{cell }}^{s}=-2.71 V$

(b) $2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} E   _{\text{cell }}^{s}-1.23 V$

(d) $Cl^{-}(aq) \longrightarrow \frac{1}{2} Cl_2(g)+e^{-} ; E   _{\text{cell }}^{s}=1.36 V$

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**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).

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**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.

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**19** $ E   _{\text{cell }}^{s}$ for some half cell reactions are given below. On the basis of these mark the correct answer.

(a) $H^{+}(aq)+e^{-} \longrightarrow \frac{1}{2} H_2(g) ; \quad \quad E   _{\text{cell }}^{s}=0.00 V$

(b) $2 H_2 O(l) \longrightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; E   _{\text{cell }}^{s}=1.23 V$

(a) In dilute sulphuric acid solution, hydrogen will be reduced at cathode

(b) In concentrated sulphuric acid solution, water will be oxidised at anode

(d) In dilute sulphuric acid solution, $SO_4^{2-}$ ion will be oxidised to tetrathionate ion at anode

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**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.

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**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

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**22** $ \Lambda_m^{\circ} H_2 0$ is equal to.......

(a) $\Lambda   _{m(HCl)}^{\circ}+\Lambda _{m(NaOH)}^{\circ}-\Lambda _{m(NaCl)}^{\circ}$

(b) $\Lambda   _{m(HNO_3)}^{\circ}+\Lambda _{m(NaNO_3)}^{\circ}-\Lambda _{m(NaOH)}^{\circ}$

(d) $\Lambda   _{m(NH_4 OH)}^{\circ}+\Lambda _{m(HCl)}^{\circ}-\Lambda _{m(NH_4 Cl)}^{\circ}$

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**Explanation**

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.

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**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.

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**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.

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**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}
$

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**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.

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**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.

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**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.

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**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 $B=-$

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**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}
$

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**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.

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**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.

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**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.

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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).

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**54.** Match the items of Column I and Column II.

|  | Column I | |Column II |
| :---: | :---: | :---: |:---: |
| A. | Lechlanche cell | 1. |Cell reaction $2 H_2+O_2 \longrightarrow 2 H_2 O$ |
| B. | $Ni-Cd$ cell | 2.| Does not involve any ion in solution and is used in hearing aids. |
| C. | Fuel cell | 3.| Rechargeable |
| D. | Mercury cell | 4.| Reaction at anode, $Zn \longrightarrow Zn^{2+}+2 e^{-}$  |
|  | | |  5. Converts energy of combustion into electrical energy |

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**55.** Match the items of Column I and Column II on the basis of data given below

$
\begin{aligned}
E   _{F_2 / F^{-}}^{s} & =2.87 V, E _{Li^{+} / Li}^{s}=-3.5 V \\\\\\
E   _{Au^{3+} / Au}^{s} & =1.4 V, E _{Br_2 / Br^{-}}^{s}=1.09 V
\end{aligned}
$

|  | Column I | |Column II |  
| :--- | :--- | :--- | :--- |
| A. | $F_2$ | 1. | Metal is the strongest reducing agent |
| B. | $Li^{\text{2. }}$ | 2. | Metal ion which is the weakest oxidising agent |
| C. | $Au^{3+}$ | 3. | Non-metal which is the best oxidising agent |

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<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

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<div style='float: left; width:100%; text-align: left;font-size:22px'>
 
 **Assertion and Reason**

In the following questions a statement of assertion (A) followed by a statement of reason ( $R$ ) is given. Choose the correct answer out of the following choices.
 
(a) Both assertion and reason are true and the reason is the correct explanation of assertion.

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

(d) Both assertion and reason are false.

(e) **Assertion** is false but reason is true.

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<footer> <font size = "4">  MCQ (Single correct option)  $\rarr$  MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$ <span style="color:blue">Assertion and Reason</span> $\rarr$ Long answer type questions  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

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**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 $E   _{\text{cathode }}>E _{\text{anode }}$.

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<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Assertion and Reason** </Primary>

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**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.

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<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

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(i) Redraw the diagram to show the direction of electron flow.

(ii) Is silver plate the anode or cathode?

(iii) What will happen if salt bridge is removed?

(iv) When will the cell stop functioning?

(v) How will concentration of $Zn^{2+}$ ions and $Ag^{+}$ions be affected when the cell functions?

(vi) How will the concentration of $Zn^{2+}$ ions and $Ag^{+}$ions be affected after the cell becomes 'dead'?

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<footer> <font size = "4">  MCQ (Single correct option)  $\rarr$  MCQ (More than one correct questions) $\rarr$ Short answer type questions $\rarr$ Matching the columns $\rarr$ Assertion and Reason $\rarr$ <span style="color:blue">Long answer type questions</span>  </font></footer>

<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

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(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.

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<Success style="float:center; font-size: 25px; text-align:center;"><b> **ELECTROCHEMISTRY**</b></Success>

#### <Primary style ="float:center; font-size: 23px; text-align:center;"> **Long answer type questions** </Primary>

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**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.

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