Electrochemistry

The correct order of negative standard potential value of $\mathrm{Li}, \mathrm{Na}, \mathrm{K}$ is $\mathrm{Li}>\mathrm{K}>\mathrm{Na}$

Stronger the reducing agent, more will be the negative standard electrode potential.

The balanced chemical equation is,

$ M(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow M^{2+}(a q)+2 \mathrm{Ag}(s) $

From this $n=2$

Using the Nernst equation,

$ E^{\circ}=\frac{2303 R T}{n F} \log k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Substituting the values,

$ \begin{aligned} & E^{\circ}=\frac{0.06 \mathrm{~V}}{2} \log \left(10^{15}\right) \\ & E^{\circ}=0.03 \mathrm{~V} \times 15=0.45 \mathrm{~V} \end{aligned} $

Given,

Current $I=0.5 \mathrm{~A}$

Time $t=96.5 \mathrm{~s}$

$ Q=i \times t=0.5 \times 96.5=48.25 \mathrm{C} $

2. Aluminium at cathode

$ \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al} $

Equivalents of $\mathrm{Al}=\frac{Q}{F}=\frac{48.25}{96500}$

$ =5 \times 10^{-4} $

Mass $=$ Equivalent weight × Equivalents

$ =9 \times 5 \times 10^{-4}=4.5 \mathrm{mg} $

3. Chlorine at anode

$ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 e^{-} $

Moles $\mathrm{Cl}_2=\frac{\left(5 \times 10^{-4}\right)}{2}$

$ =2.5 \times 10^{-4} $

Volume at STP $=2.5 \times 10^{-4} \times 22.4 \times 1000$

$ =5.6 \mathrm{~mL} $

(a) Calculate the degree of dissociation ( $\alpha$ ).

For a weak acid, $K_a=C \alpha^2$ (assuming $1-\alpha \approx 1$ ).

Given, $K_a=1.8 \times 10^{-5}$ and $C=0.01 \mathrm{M}$.

$ \begin{aligned} \alpha^2 & =K_a / C=\left(1.8 \times 10^{-5}\right) / 0.01=1.8 \times 10^{-3} \\ \alpha & =\sqrt{1.8 \times 10^{-3}} \approx 0.0424 \end{aligned} $

(b) Calculate the molar conductivity ( $\Lambda_m$ )

The relationship is $\Lambda_m=\alpha \Lambda_m^0$.

Given $\Lambda_m^{\circ}=390 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$.

$ \begin{aligned} & \Lambda_m=0.0424 \times 390 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \approx 16.54 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} . \end{aligned} $

Statement given in option (c) is incorrect about Castner-Kellner cell process. It's correct form is, mercury acts as cathode and carbon rod acts as anode.

The Castner-Kellner cell process is an electrolytic method for the industrial production of sodium hydroxide, chlorine, and hydrogen from brine. Let's evaluate each statement:

• Option A: sodium hydroxide is prepared.

  In the Castner-Kellner cell, sodium ions are reduced at the mercury cathode to form sodium amalgam. This amalgam is then reacted with water in a separate decomposer unit to produce sodium hydroxide (NaOH) and hydrogen gas. So, this statement is correct.

• Option B: brine solution is the electrolyte.

  The cell uses an aqueous solution of sodium chloride (NaCl), which is commonly known as brine, as the electrolyte. So, this statement is correct.

• Option C: mercury acts as anode and carbon rod acts as cathode.

  In the Castner-Kellner cell:

  • Anode: Graphite (carbon) rods are used as the anode, where chloride ions are oxidized to chlorine gas (2Cl⁻(aq) → Cl₂(g) + 2e⁻).

  • Cathode: A flowing layer of mercury acts as the cathode, where sodium ions are reduced to form sodium amalgam (Na⁺(aq) + e⁻ → Na(in Hg)).

  Therefore, the statement that mercury acts as the anode and carbon rod acts as the cathode is incorrect. The roles are reversed.

• Option D: chlorine gas liberates at anode.

  As mentioned above, at the graphite anode, chloride ions are oxidized to produce chlorine gas. So, this statement is correct.

The incorrect statement is C.

The final answer is $\boxed{C}$

$ \begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & =(-0.45)+(0.75)=0.30 \mathrm{~V} \end{aligned} $

For the reaction:

$ \begin{aligned} & 2 \mathrm{Cr}+3 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{Fe} \\ & Q=\frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Fe}^{2+}\right]^3}=\frac{(0.1)^2}{(0.001)^2}=10^4 \\ & E_{\text {cell }}=0.30-\frac{0.059}{6} \log 10^4 \\ &=0.030-\frac{0.059}{6} \times 4 \\ & \approx 0.2607 \mathrm{~V} \end{aligned} $

Gibb's free energy,

$ \begin{aligned} \Delta G^{\circ} & =-n F E^{\circ} \\ & =6 \times 96500 \times 0.2607 \times 10^{-3} \\ & =-150.9 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Reaction at anode is,

$ 2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} $

This produce gas $=\mathrm{O}_2(X)$

Reaction at cathode

$ \mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}(s) $

Metal deposited = $\mathrm{Cu}(Y)$

The cell reaction is,

At anode $(\mathrm{Zn}), \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-}$

Cathode $\left(\mathrm{H}^{+}\right), 2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2$

Using Nernst equation,

$ \begin{aligned} & E_{\text {cell }}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}-\frac{2.303 R T}{F} \log Q \\ & 0.28=+0.76-\frac{0.06}{2} \log \left(\frac{0.01}{\left[\mathrm{H}^{+}\right]^2}\right) \end{aligned} $

Solving this,

$ \begin{aligned} {\left[\mathrm{H}^{+}\right] } & =10^{-9} \\ \mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=9 \end{aligned} $

Given, Mass of copper $ =0.592 \mathrm{~g} $, time $ =60 $ minutes, current $ =0.5 \mathrm{~A} $ Total charge passes through solution,

$ \begin{array}{l} Q=I \times t \\\\ Q=0.5 \times 60 \times 60=1800 \mathrm{C} \end{array} $

Number of moles of electrons transferred.

$ \begin{array}{l} n=\frac{Q}{F}=\frac{1800 \mathrm{C}}{96500 \mathrm{C} / \mathrm{mol}} \\\\ \Rightarrow \quad n=0.01865 \mathrm{~mol} \end{array} $

$ \mathrm{Cu}^{2+} $ requires 2 moles of electron to deposit 1 mole of copper

$ \begin{array}{c} \text { Moles of copper }=\frac{n}{2}=\frac{0.01865}{2} \\\\ =0.009325 \mathrm{~mol} \end{array} $

Mass of copper deposited

$ \begin{array}{l} =0.009325 \times 63.55 \\\\ =0.592 \mathrm{~g} \end{array} $

Electrochemical equivalent is given by

$ \frac{\text { Mass of copper deposited }}{\text { Charged passed }}=\frac{0.592}{180} $

$ =3.3 \times 10^{-4} \mathrm{~g} / \mathrm{C} $

The standard electrode potentials for the $\mathrm{Li}^{+} / \mathrm{Li}$ and $\mathrm{Na}^{+} / \mathrm{Na}$ half-cells are as follows:

$\mathrm{Li}^{+} / \mathrm{Li}$: $E^{\circ} = -3.04 \, \mathrm{V}$

$\mathrm{Na}^{+} / \mathrm{Na}$: $E^{\circ} = -2.714 \, \mathrm{V}$

Therefore, the correct option is:

Option A

$ -3.04, -2.714 $

These values indicate the tendency of lithium and sodium to lose electrons and form cations, with lithium being more reactive in this context due to its more negative electrode potential.

Statement $ I $ is correct but statement II is incorrect. Correct form of Statement II is

At cathode (Reduction take place)

$ 2 \mathrm{Cu}^{2+}+4 e^{-} \xrightarrow{}2 \mathrm{Cu} $

At anode (oxidation)

$ 4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-} $

Hence, oxygen is liberated at anode not cathode.

Given,

Molar conductivity

$ =124 \times 10^{-4} \mathrm{Sm}^{2} / \mathrm{mol} $

Molarity $ =0.02 \mathrm{M} $,

Cell constant $ =129 \mathrm{~m}^{-1} $

Molar conductivity is given by

$ \Lambda_{\mathrm{m}}=\frac{1000 \times \kappa}{M} $

where $ \kappa $ is conductivity

$ \mathrm{K}=\frac{\Lambda_{\mathrm{m}} \times M}{1000} $

Substitute the vaues in Eq. (i),

$ \kappa=\frac{124 \times 10^{-4} \times 0.02}{1000} \Rightarrow \kappa=0.248 \ldots $

Conductivity, k is given by

$ \begin{array}{c} \mathrm{K}=\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \text { or Resistance }=\frac{\text { Cell constant }}{\mathrm{K}} \\\\ R=\frac{129}{0.248} \text { or } R=520 \Omega \end{array} $

Using Kohlrausch's law we get, (for formic acid)

$ \begin{aligned} & \left.\lambda_{(\mathrm{HCOOH})}^{\circ}=\lambda_{\left(\mathrm{H}^{+}\right)}^{\circ}+\lambda_{(\mathrm{HCOO})}^{\circ}\right) \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=349.6+54.6 \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=404.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Degree of dissociation is given by

$ \alpha=\frac{\lambda_{(\mathrm{HCOOH})}}{\lambda_{(\mathrm{HCOOH})}^{\circ}} $

[where, $\alpha=0.11$ (as $11 \%$ given)]

$ \begin{aligned} \lambda_{(\mathrm{HCOOH})} & =\text { Molar conductivity of } \mathrm{HCOOH} \\\\ \lambda_{(\mathrm{HCOOH})} & =\alpha \times \lambda_{(\mathrm{HCOOH})}^{\circ} \\\\ & =0.11 \times 404: 2 \\\\ & =44.46 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Correct Statements

To evaluate the given statements, we analyze each one:

Statement (A): The potential of the hydrogen electrode in a solution of $\text{pH} = 10$ at 298 K.

The potential of a hydrogen electrode can be calculated using the Nernst equation for the hydrogen electrode:

$ E = E^\ominus - \frac{0.0591}{n} \log [\text{H}^+] $

At standard conditions $E^\ominus = 0$, and $[\text{H}^+]$ is calculated from the pH:

$ \text{pH} = 10 \implies [\text{H}^+] = 10^{-10} $

Substitute into the Nernst equation:

$ E = -0.0591 \times \log(10^{-10}) = -0.59 \, \text{V} $

Statement A is correct.

Statement (B): Limiting molar conductivity of $\text{Ca}^{2+}$ and $\text{Cl}^{-}$ are 119 and 76 S cm$^2$ mol$^{-1}$ respectively.

The limiting molar conductivity of $\text{CaCl}_2$ is the sum of the molar conductivities of $\text{Ca}^{2+}$ and two $\text{Cl}^{-}$ ions:

$ \lambda^\circ_{\mathrm{CaCl}_2} = \lambda^\circ_{\mathrm{Ca}^{2+}} + 2\lambda^\circ_{\mathrm{Cl}^-} $

Substitute the given values:

$ \lambda^\circ_{\mathrm{CaCl}_2} = 119 + 2(76) = 119 + 152 = 271 \, \text{S cm}^2 \text{ mol}^{-1} $

The statement given as 195 S cm$^2$ mol$^{-1}$ is incorrect.

Statement (C): Relationship between $K_C$ and $E_{\text{cell}}^{\ominus}$.

The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation:

$ E_{\text{cell}}^{\ominus} = \frac{RT}{nF} \ln K_C $

Converting natural log to base-10 log:

$ E_{\text{cell}}^{\ominus} = \frac{2.303 RT}{nF} \log K_C $

The statement is correctly stated.

Conclusion

Based on the analysis, the correct statements are A and C.

Option C: A, C only is the correct answer.

The half-cell reaction will be

$ \begin{aligned} \mathrm{Fe}^{3+}(0.02 \mathrm{M})+e^{-} \longrightarrow & \mathrm{Fe}^{2+}(2.0 \mathrm{M}) ; \\ & E^{\circ}{ }_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=0.771 \mathrm{~V} \end{aligned} $

According to Nernst's equation,

$ E=E_{\mathrm{Cell}}^{\circ}-\frac{2303 R T}{n F} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} $

[Number of electrons involved in reduction, $n=1$ ]

$ \begin{gathered} E=0.771 \mathrm{~V}-\frac{2303 R T}{F} \log \frac{2}{0.02} \\ =0.771 \mathrm{~V}-0.059 \log 100 \\ =0.771-0.118=0.653 \mathrm{~V} \end{gathered} $

Hence, reduction potential of half-cell is 0.653 V .

Current $=15 \mathrm{~A}$

$ \begin{aligned} & \text { Time }=45 \times 60=2700 \mathrm{~s} \\ & \text { Since, } Q=I \times t \\ & \\ & \\ & Q=15 \times 2700=40,500 \mathrm{C} \end{aligned} $

Now, for reduction of chlorine gas

$ \mathrm{Cl}_2+2 e^{\ominus} \longrightarrow 2 \mathrm{Cl}^{\ominus} $

or, 2 mole of $e^{\ominus}$ one required for 1 mole of $\mathrm{Cl}_2$

or, $2 \times 96500 \mathrm{C}$ of is required for 22.4 L of $\mathrm{Cl}_2$ electricity.

$\therefore 40,500 \mathrm{C}$ will require ' $x$ '.

$ \Rightarrow \quad x=\frac{40500 \times 22.4}{2 \times 96500}=4.7 \mathrm{~L} $

$ \begin{aligned}Given,\,\, & E_{A^{+} / A}^{\circ}=-0.5 \mathrm{~V} \\ & E_{B^{+} / B}^{\circ}=0.5 \mathrm{~V} \end{aligned} $

For reaction,

$ \begin{aligned} A(s)+B^{+}(a q) & \rightleftharpoons A^{+}(a q)+B(s) \\ E_{\text {cell }}^{\circ} & =+0.5-(-0.5)=+1.0 \\ E_{\text {cell }}^{\circ} & =\frac{2.303 R T}{n F} \log K_C \quad[\because n=1] \\ 1 & =\frac{0.06}{1} \log K_C \\ & \quad\left[\text { Given, } \frac{2.303 R T}{F}=0.06 \mathrm{~V}\right] \\ \frac{1}{0.06} & =\log K_C \text { or } \frac{100}{6}=\log K_C \end{aligned} $

Degree of dissociation $(\alpha)=$

Molar conductivity at $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$

Molar conductivity at infinity dilution

$ \begin{aligned} 0.02 & =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{\Lambda_{\mathrm{m}}}{400 \mathrm{Scm}^2 \mathrm{~mol}^{-1}} \\ \Lambda_{\mathrm{m}} & =8 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \end{aligned} $

Also,

$ \begin{aligned} \Lambda_{\mathrm{m}} & =\frac{\text { Conductivity }(\kappa) \times 1000}{\text { Concentration of solution }(C)} \\ 8 & =\frac{\kappa \times 1000}{0.1} \end{aligned} $

Conductivity, $\kappa=8 \times 10^{-4} \mathrm{Scm}^{-1}$

To determine the degree of dissociation of acetic acid, follow these steps:

Concentration of Acetic Acid:

$ C = 0.01 \, \text{mol dm}^{-3} = 10 \, \text{mol m}^{-3} $

Conductivity ($\kappa$):

$ \kappa = 19.5 \times 10^{-5} \, \text{S cm}^{-1} = 19.5 \times 10^{-3} \, \text{S m}^{-1} $

Limiting Molar Conductivity ($\lambda_m^\circ$):

$ \lambda_m^\circ = 390 \, \text{S cm}^2 \text{mol}^{-1} = 390 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} $

Calculate Molar Conductivity ($\lambda_m$):

$ \lambda_m = \frac{\kappa}{C} = \frac{19.5 \times 10^{-3}}{10} = 19.5 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} $

Degree of Dissociation ($\alpha$):

$ \alpha = \frac{\lambda_m}{\lambda_m^\circ} = \frac{19.5 \times 10^{-4}}{390 \times 10^{-4}} = 0.05 = 5 \times 10^{-2} $

Therefore, the degree of dissociation of acetic acid is $5 \times 10^{-2}$.

Let's analyze the problem step by step.

The chlorine electrode involves the half‐reaction

$Cl_2(g) + 2e^- \rightarrow 2Cl^-.$

Its potential at 25°C is given by the Nernst equation

$E = E^\circ - \frac{0.059}{2}\log \frac{[Cl^-]^2}{P_{Cl_2}},$

where $P_{Cl_2}$ is the partial pressure of chlorine gas (in atm) and $[Cl^-]$ is the activity (or concentration) of the chloride ion.

Note the following:

For fixed $P_{Cl_2}$ (usually taken as 1 atm), the electrode potential $E$ is a decreasing function of the logarithm of $[Cl^-]^2$.

Therefore, to get the maximum electrode potential, the ratio $\frac{[Cl^-]^2}{P_{Cl_2}}$ must be as small as possible—that is, $[Cl^-]$ must be minimized.

In practice the “chlorine electrode” is prepared in a solution with no added chloride. However, when chlorine gas is bubbled into pure water, it undergoes the following hydrolysis:

$Cl_2 + H_2O \leftrightarrow HCl + HOCl.$

Since HCl is a strong acid, it dissociates completely:

$HCl \rightarrow H^+ + Cl^-.$

In the absence of any added chloride, the only source of $[Cl^-]$ in the solution is from this hydrolysis. Experiments have shown that under these conditions (i.e. when the solution is free of extrinsic chloride) the concentration of chloride ion attains a minimum value.

It turns out that this minimum (and hence the maximum electrode potential) corresponds to a chloride concentration of approximately

$2.5 \times 10^{-3}\; \text{mol L}^{-1}.$

Thus, among the options given, the maximum potential is achieved when

$X = 2.5 \times 10^{-3}\; \text{mol L}^{-1}.$

So, the correct answer is Option A.

$ \begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} & \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+} \\ E_{\mathrm{Ce}^{4+}}^{\circ} / \mathrm{Ce}^{3+} & =1.6 \mathrm{~V} \\ E_{\mathrm{Fe}^{3+}}^{\circ} / \mathrm{Fe}^{2+} & =0.76 \mathrm{~V} \\ E_{\text {cell }}^{\circ} & =E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \end{aligned} $

$ \begin{aligned} \mathrm{Fe}^{3+}+e^{-} \longrightarrow & \mathrm{Fe}^{2+} \\ & \left(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.76 \mathrm{~V}\right) \\ \mathrm{Ce}^{4+}+e^{-} \longrightarrow & \mathrm{Ce}^{3+} \\ & \left(E_{\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}}^{\circ}=1.6 \mathrm{~V}\right) \end{aligned} $

$\mathrm{Fe}^{3+}$ is oxidising $\mathrm{Ce}^{3+}$, i.e. it is itself getting reduced.

Also,

$ \begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {red }}^{\circ}-E_{\text {oxidation }}^{\circ} \\ & =0.76-1.6=-0.84 \mathrm{~V} \end{aligned} $

At cathode,

$ 2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} $

At anode, $2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+4 e^{-}+\mathrm{O}_2$

At anode,

$ \begin{aligned} E_{\mathrm{O}_2} & =\frac{32}{4}=8, W_{\mathrm{O}_2}=\frac{E I t}{96500} \\ W_{\mathrm{O}_2} & =\frac{32 \times 3 \times 10 \times 60}{4 \times 96500} \\ W_{\mathrm{O}_2} & =0.1492 \mathrm{~g} \end{aligned} $

At STP, volume of $\mathrm{O}_2=\frac{0.1492 \times 22.4}{32}$

$ =0.104 \mathrm{~L} . $

Variation of molar conductivity ( $\lambda_{\mathrm{m}}$ ) of acetic acid with concentration vary as follows.

For weak electrolyte, molar conductivity is indirectly proportional to concentration.

So, in case of acetic acid, $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$, the proportion of dissociated ions is high at low concentration, hence, its molar conductivity is high.

Since, same quantity of electricity is passed through both silver ( Ag ) and gold ( Au ) solution.

So,  $ \frac{\text { Mass of } \mathrm{Au} \text { deposited }}{\text { Mass of } \mathrm{Ag} \text { deposited }}=\frac{\text { equivalent mass of } \mathrm{Au}}{\text { equivalent mass of } \mathrm{Ag}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} \text { Equivalent mass of } \mathrm{Au} & =\frac{\text { molecular mass of } \mathrm{Au}}{\text { valency of } \mathrm{Au}} \\ & =\frac{197}{\text { valency of } \mathrm{Au}} \\ \text { Equivalent mass of } \mathrm{Ag} & =\frac{\text { molecular mass of } \mathrm{Ag}}{\text { valency of } \mathrm{Ag}}=\frac{107.9}{1} \end{aligned} $

From Eq. (i)

$ \frac{1.31}{2.15}=\frac{197}{\text { valency of } \mathrm{Au}} \times \frac{1}{107.9} $

$\therefore \quad$ Valency of $\mathrm{Au}=\frac{197}{1.31} \times \frac{2.15}{107.9}=2.9 \approx 3$

Hence, valency of $\mathrm{Au}=3$.

$ \text { Given, } E^{\circ}{ }_{\mathrm{Mn}^{7+}} /_ {\mathrm{Mn}^{2+}}=1.51 \mathrm{~V} $

$ E^{\circ} _{\mathrm{Mn}^{4+}} / _{\mathrm{Mn}^{2+}}=1.23 \mathrm{~V} $

$ \text { Latimer diagram, is as follows } $

$ \begin{aligned} &\text { }\\ &\begin{aligned}So,\,\, n_1 E_1 & =n_2 E_2+n_3 E_3 \\ 5 \times 1.51 & =2 \times 1.23+3 \times x \\ 7.55 & =2.46+3 x \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 3 x=7.55-2.46 \\ & \,\,\,\,\,\,\,\,\,\, \quad 3 x=5.09 \\ & \Rightarrow \quad x=\frac{5.09}{3} \Rightarrow x=1.69 \end{aligned}\\ &\text { So, } \quad x=+1.7 \mathrm{~V} \text {. } \end{aligned} $

Current $=1.2 \mathrm{~A}$

Time $=40$ minutes $=2400 \mathrm{~s}$

Mass of copper deposited $=0.96 \mathrm{~g}$

Coulombs of charge $=$ charge × time

$ \begin{aligned} & =1.2 \times 2400 \\ & =2880 \mathrm{C} \end{aligned} $

Moles of electrons in $2880 \mathrm{C}=\frac{2880}{96500}$

$ \begin{aligned} & =0.02984 \\ & =2.98 \times 10^{-2} \end{aligned} $

Equivalent weight of Cu

$ \begin{aligned} & =\frac{\text { Mass of Cu deposited }}{\text { Moles of electrons }} \\ & =\frac{0.96}{2.98 \times 10^{-2}}=31.75 \end{aligned} $

Consider the reduction reaction

$ M^{n+}+n e^{-} \longrightarrow M $

According to Faradays law,

$ W=\frac{M}{n} \times \frac{i \times t}{F}=\frac{M}{n} \times \frac{Q}{F} $

If $W$ is the weight of substance deposited.

Since $W$ is given as 1 g equivalent, which is equal to $M / n$. Therefore, $Q=F$.

$1 F=96500$ coulomb of charge = charge on 1 mole of electrons. Therefore, to deposit 1 g equivalent of substance 1 mole of electrons is needed.

$\because E=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]}$

Given, $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$

and, $\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}$

∴ For $80 \%$ of $\mathrm{Fe}^{2+}$

$ \begin{array}{rl} \mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+}+e^{-} \\ \text {Initial }=80 & 20 \\ \text { Final }=20 & 80 \end{array} $

$ \begin{aligned} \because \quad E & =0.77-0.059 \log \frac{80}{20} \\ & =0.77-0.05 \log 4 \\ \therefore \quad E & =0.77-0.0592 \log 2 \\ & =0.77-0.06=0.77-0.0354 \\ & =0.734 \end{aligned} $

Hence, option (c) is the correct answer.

The galvanic cell can be represented as :

$ \begin{aligned} & \mathrm{Mg}\left|\mathrm{Mg}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu} \\ & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ &=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ} \\ & \Rightarrow E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}-E_{\mathrm{Cell}}^{\circ} \end{aligned} $

(Reduction potential)

$ \begin{aligned} & =0.34-2.71 \mathrm{~V}=-2.37 \mathrm{~V} \\ & =(0.34-2.71) \mathrm{V}=-2.37 \mathrm{~V} \end{aligned} $

$\because \Delta G=(-) W=n F E_{\text {cell }}^{\circ}$ where, $\Delta G=$ free energy, $W=$ maximum work done

$F=96500 \mathrm{C}, \quad E=$ cell potential and $E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$

$n=$ number of electrons envolved

Given, $\quad E_{\text {cathode }}^{\circ}=0.8 \mathrm{~V}$

$E_{\text {anode }}^{\circ}=(-) 1.7 \mathrm{~V}$

$\therefore \quad E_{\text {cell }}^{\circ}=0.8-[(-) 1.7]=2.5 \mathrm{~V}$

$ \begin{aligned} \therefore \quad W_{\text {maximum }} & =\frac{2 \times 96500 \times 2.5}{1000}(\text { in } \mathrm{kJ} / \mathrm{mol}) \\ & =482.5 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Hence, option (b) is the correct answer.

Electrode having higher reduction potential has to be cathode.

Hence, silver electrode act as a cathode. Because reduction potential of silver electrode $\left(E^{\circ}\right)$ is +80 V . Which is more than given reduction potential of $\mathrm{Cu}(+34 \mathrm{~V})$.

Hence, $E_{\text {cell }}^{\circ}$ will determine as:

$ \begin{aligned} E_{\text {cell }}^{\circ} & =E_{(\text {Reduction }) \text { cathode }}^{\circ}-E_{(\text {Reduction }) \text { anode }}^{\circ} \\ & =+80-(+0.34)=+0.46 \mathrm{~V} \end{aligned} $