Electrochemistry

(1) If $\mathrm{C}_1$ is at anode ⇒ cell reaction

$ \begin{aligned} & \mathrm{M}^{+}\left(\mathrm{C}_2\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_1\right) \\ & \mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_1}{\mathrm{C}_2} \\ & \therefore \mathrm{E}_{\text {cell }}>0 \Rightarrow \mathrm{C}_1<\mathrm{C}_2 \end{aligned} $

(2) If $\mathrm{C}_1$ is at cathode

$ \begin{aligned} & \mathrm{M}^{+}\left(\mathrm{C}_1\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_2\right) \\ & \mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_2}{\mathrm{C}_1}>0 \\ & \mathrm{C}_2<\mathrm{C}_1 \end{aligned} $

$\begin{aligned} & \mathrm{Cl}_{\mathrm{aq}}^{-}+\mathrm{Ag}_{\mathrm{S})}+\mathrm{Fe}_{\mathrm{aq}}^{+3} \rightarrow \mathrm{Fe}_{\mathrm{aq}}^{+2}+\mathrm{AgCl}_{(\mathrm{S})} \\ & \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Cl}^{-}\right]\left[\mathrm{Fe}^{3+}\right]}\end{aligned}$

Reducing agent is the species that gets oxidised (donates electrons). In NCERT terms:

• More negative $E^\circ_{\text{red}}$ for a couple $\Rightarrow$ the reduced form is a stronger reducing agent (it will be more easily oxidised).

• Equivalently, higher $E^\circ_{\text{ox}}$ $\Rightarrow$ stronger reducing agent.

Now write oxidation potentials for the given reducing species:

1) From $ \mathrm{Al}^{3+}+3e^- \rightarrow \mathrm{Al}(s),\ E^\circ_{\text{red}}=-1.66\text{ V}$

$ \mathrm{Al}(s)\rightarrow \mathrm{Al}^{3+}+3e^- ,\quad E^\circ_{\text{ox}}=+1.66\text{ V} $

2) From $ \mathrm{Cr}^{3+}+3e^- \rightarrow \mathrm{Cr}(s),\ E^\circ_{\text{red}}=-0.74\text{ V}$

$ \mathrm{Cr}(s)\rightarrow \mathrm{Cr}^{3+}+3e^- ,\quad E^\circ_{\text{ox}}=+0.74\text{ V} $

3) From $ \mathrm{Fe}^{3+}+e^- \rightarrow \mathrm{Fe}^{2+},\ E^\circ_{\text{red}}=+0.77\text{ V}$

$ \mathrm{Fe}^{2+}\rightarrow \mathrm{Fe}^{3+}+e^- ,\quad E^\circ_{\text{ox}}=-0.77\text{ V} $

4) From $ \mathrm{Co}^{3+}+e^- \rightarrow \mathrm{Co}^{2+},\ E^\circ_{\text{red}}=+1.81\text{ V}$

$ \mathrm{Co}^{2+}\rightarrow \mathrm{Co}^{3+}+e^- ,\quad E^\circ_{\text{ox}}=-1.81\text{ V} $

Now arrange reducing strength (decreasing $E^\circ_{\text{ox}}$):

$ +1.66 > +0.74 > -0.77 > -1.81 $

So, tendency to act as reducing agent decreases as:

$ \boxed{\mathrm{Al} > \mathrm{Cr} > \mathrm{Fe}^{2+} > \mathrm{Co}^{2+}} $

Correct option: D

Standard cell potential ($\mathrm{E}_{\text{cell}}^0$) does not change as time passes.

The relation between molar conductivity and concentration for strong electrolytes is given by

$\Lambda_M = \Lambda_M^0 - b\sqrt{c}$

For two concentrations $ c_1 $ and $ c_2 $, the equation becomes:

$\Lambda_{M,1} - \Lambda_{M,2} = b(\sqrt{c_2} - \sqrt{c_1})$

Therefore,

$b = \frac{\Lambda_{M,1} - \Lambda_{M,2}}{\sqrt{c_2} - \sqrt{c_1}} = \frac{\Lambda_{M,1} - \Lambda_{M,2}}{0.1}$

For NaNO₃:

$b = \frac{111 - 101}{0.1} = 100$

So the limiting molar conductivity is:

$[\Lambda_M^0]_{\mathrm{NaNO_3}} = \Lambda_M + b\sqrt{c} = 111 + 100(0.1) = 121\, \mathrm{S\,cm^2/mol}$

For NaCl:

$b = \frac{117 - 107}{0.1} = 100$

$[\Lambda_M^0]_{\mathrm{NaCl}} = 117 + 100(0.1) = 127\, \mathrm{S\,cm^2/mol}$

For AgNO₃:

$b = \frac{125 - 116}{0.1} = 90$

$[\Lambda_M^0]_{\mathrm{AgNO_3}} = 125 + 90(0.1) = 134\, \mathrm{S\,cm^2/mol}$

Finding limiting molar conductivity of AgCl:

Using Kohlrausch’s law of independent migration of ions:

$[\Lambda_M^0]_{\mathrm{AgCl}} = [\Lambda_M^0]_{\mathrm{AgNO_3}} + [\Lambda_M^0]_{\mathrm{NaCl}} - [\Lambda_M^0]_{\mathrm{NaNO_3}}$

$= 134 + 127 - 121 = 140\, \mathrm{S\,cm^2/mol}$

Now, to find solubility (X):

The conductivity of the saturated solution is $ k = 1.40 \times 10^{-6}\, \mathrm{S\,cm^{-1}} $.

Since the electrolytic solution is very dilute,

$\Lambda_M = \Lambda_M^0 = \frac{k \times 1000}{c}$

So,

$c = \frac{k \times 1000}{\Lambda_M^0} = \frac{1.4\times10^{-6}\times10^3}{140} = 10^{-5}\, \mathrm{M}$

Thus, solubility $ X = 10^{-5} \, \mathrm{mol\,L^{-1}} $.

Finally,

$\log(X^{-1}) = \log(10^{-5})^{-1} = 5$

(i) $\mathrm{Fe}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) \ldots \ldots . . \Delta \mathrm{G}_1^{\circ}$

(ii) $\mathrm{Fe}^{+3}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) \ldots \ldots . . \Delta \mathrm{G}_2{ }^{\circ}$

(iii) $\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2}(\mathrm{aq}) \ldots \ldots . . \Delta \mathrm{G}_3^{\circ}$

$ \begin{aligned} & \text { iii }=\text { ii }-\mathrm{i} \\ & -1 \times \mathrm{F}\times \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=-3 \times \mathrm{F} \times \mathrm{y}-(-2 \times \mathrm{F} \times \mathrm{x}) \\ & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=3 \mathrm{y}-2 \mathrm{x} \end{aligned} $

Option (B) is correct

$ \begin{aligned} &\text { }\\ &\begin{aligned}(a)\,\, Anode:\,\, & \left(R{e d 1} \longrightarrow O_{x_1}^{n^{+}}+u_1 e^{-}\right) \cdot n_2 \\ & \left(Ox_{2}+n_2 e^{-} \longrightarrow \operatorname{Red}_2^{n_2^{-}}\right) \cdot n_1 \end{aligned} \end{aligned} $

To add the anode and cathode half-reactions, the number of electrons must be the same on both sides.

So, we multiply the anode reaction by $n_2$ and the cathode reaction by $n_1$, so that both have $n_1 n_2$ electrons.

$ \begin{aligned} &\begin{aligned} & n_2 \operatorname{Red}_1 \longrightarrow n_2 \mathrm{Ox}_1^{n+}+n_2 n_1 e^{-} \\ & n_1 \mathrm{Ox}_2+n_1 n_2 e^{-} \longrightarrow n_1 \operatorname{Red}_2^{n_2-} \end{aligned}\\ &\text { Overall ren: } n_2 R{e d}+n_1 O x_2 \longrightarrow n_2 O x^{n+}+n_1 R{e d_ 2}^{n_2-} \end{aligned} $

Now, when we add these two equations, the $n_1 n_2 e^{-}$ on the product side of the anode reaction cancels with the $n_1 n_2 e^{-}$ on the reactant side of the cathode reaction.

(b) The $e^{-}$produced at the anode are consumed at the cathode, that's why $e^{-}$do not appear in the overall reaction.

This statement is correct because electrons cancel during addition of the two half-reactions.

(c) We know,

$ \begin{aligned} & E = E^{o}-\frac{R T}{n f} \log Q \\ & E-E^{\circ}=\frac{-R T}{n f} \log Q \\ & \frac{E-E^{\circ}}{R T / f}=-\frac{1}{n} \log Q \end{aligned} $

Here, $E$ changes with $\log Q$ according to the Nernst equation, and $n$ is the number of electrons transferred in the balanced redox reaction.

plotting a $y=m x+c$ graph,

$ \begin{aligned} & y=\frac{E-E^0}{R T / F} \\ & x=\log Q \\ & \text { slope }(m)=-\frac{1}{n} \\ & \text { slope } \propto \frac{1}{n} \end{aligned} $

So, the slope is $-\frac{1}{n}$. This statement is also correct.

$ \text { Where, } n=\text { no. of } e^{-} \text {transfered in redox reaction. } $

(D) The electrical work done by a galvanic cell is given by the product of the total charge transfered and the potential difference, not their ratio.

This statement is correct because electrical work is given by $W = qV$ (charge multiplied by potential difference).

$ \begin{aligned} W_{electrical} & =\text { charge × Potential difference. } \\ & =Q \times V \end{aligned} $

                $ W_e=I t \times V $

$ \mathrm{BaSO_4 (s)} \rightleftharpoons \mathrm{Ba^{2+} (aq)} + \mathrm{SO_4^{2-} (aq)} $

If the solubility of $\mathrm{BaSO_4}$ is $s \, \text{mol L}^{-1}$, then in solution:

$ [\mathrm{Ba^{2+}}] = s, \quad [\mathrm{SO_4^{2-}}] = s $

Thus, the solubility product is

$ K_{\text{sp}} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}] = s^2 $

Relation between molar conductivity and observed conductivity

By definition,

$ \kappa = \frac{\Lambda_m \, c}{1000} $

where

• $\kappa$ = conductivity in $\text{S cm}^{-1}$,

• $\Lambda_m$ = molar conductivity in $\text{S cm}^2 \text{mol}^{-1}$,

• $c$ = molar concentration in $\text{mol L}^{-1}$.

For the saturated solution of $\mathrm{BaSO_4}$,

$ \kappa = x $

and

$ c = s $

Therefore,

$ \Lambda_m^{\text{(BaSO}_4)} = \frac{1000x}{s} $

Determining $\Lambda_m^{\circ}$ for $\mathrm{BaSO_4}$

We can find $\Lambda_m^{\circ}$ for $\mathrm{BaSO_4}$ using Kohlrausch’s law of independent migration of ions.

According to the law,

$ \Lambda_m^{\circ}(\text{Electrolyte}) = \text{sum of molar ionic conductivities of its ions}. $

Now, from the data:

$ \begin{aligned} \Lambda_m^{\circ}(\mathrm{BaCl_2}) &= \lambda^{\circ}_{\mathrm{Ba^{2+}}} + 2\lambda^{\circ}_{\mathrm{Cl^-}} = x_1 \\ \Lambda_m^{\circ}(\mathrm{H_2SO_4}) &= 2\lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{SO_4^{2-}}} = x_2 \\ \Lambda_m^{\circ}(\mathrm{HCl}) &= \lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{Cl^-}} = x_3 \end{aligned} $

To get $\Lambda_m^{\circ}(\mathrm{BaSO_4}) = \lambda^{\circ}_{\mathrm{Ba^{2+}}} + \lambda^{\circ}_{\mathrm{SO_4^{2-}}}$,

we combine the above equations:

$ (x_1 + x_2 - 2x_3) = (\lambda^{\circ}_{\mathrm{Ba^{2+}}} + 2\lambda^{\circ}_{\mathrm{Cl^-}}) + (2\lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{SO_4^{2-}}}) - 2(\lambda^{\circ}_{\mathrm{H^+}} + \lambda^{\circ}_{\mathrm{Cl^-}}) $

Simplifying,

$ x_1 + x_2 - 2x_3 = \lambda^{\circ}_{\mathrm{Ba^{2+}}} + \lambda^{\circ}_{\mathrm{SO_4^{2-}}} = \Lambda_m^{\circ}(\mathrm{BaSO_4}) $

Thus,

$ \Lambda_m^{\circ}(\mathrm{BaSO_4}) = x_1 + x_2 - 2x_3 $

Equating conductivities

Since it is given that $(\Lambda_m = \Lambda_m^{\circ})$ for the saturated solution:

$ \Lambda_m^{\circ} = \frac{1000x}{s} $

$ s = \frac{1000x}{\Lambda_m^{\circ}} = \frac{1000x}{x_1 + x_2 - 2x_3} $

Expression for solubility product

$ K_{\text{sp}} = s^2 = \left(\frac{1000x}{x_1 + x_2 - 2x_3}\right)^2 = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2} $

Final Answer

$ \boxed{K_{\text{sp}} = \dfrac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}} $

Hence, Option A is correct.

For the cell,

• anode: $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^-$

• cathode: $\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}$

So the overall cell reaction is

$ \mathrm{Zn} + 2\mathrm{Ag}^+ \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{Ag} $

Now, standard cell potential:

$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $

$ E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56 \text{ V} $

Given:

$ E_{\text{cell}} = 1.60 \text{ V} $

Using Nernst equation at $298\ \text{K}$:

$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q $

Here, $n=2$.

Reaction quotient:

$ Q=\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Ag}^+]^2} $

Since $\mathrm{ZnSO}_4$ is $1$ M,

$ [\mathrm{Zn}^{2+}] = 1 $

Let

$ [\mathrm{Ag}^+] = x $

Then

$ Q=\frac{1}{x^2} $

Substitute in Nernst equation:

$ 1.60 = 1.56 - \frac{0.059}{2}\log\left(\frac{1}{x^2}\right) $

Now,

$ \log\left(\frac{1}{x^2}\right) = -2\log x $

So,

$ 1.60 = 1.56 - \frac{0.059}{2}(-2\log x) $

$ 1.60 = 1.56 + 0.059\log x $

$ 1.60 - 1.56 = 0.059\log x $

$ 0.04 = 0.059\log x $

$ \log x = \frac{0.04}{0.059} = \frac{4}{5.9} $

Hence, the correct answer is

$ \boxed{\frac{4}{5.9}} $

So, Option B is correct.

For a spontaneous electrochemical cell, the half-reaction with higher standard reduction potential undergoes reduction, and the other one undergoes oxidation.

Given half-reactions:

$ \mathrm{Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^-(aq)} \qquad E^\theta = -0.88\ \text{V} $

$ \mathrm{AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)} \qquad E^\theta = +0.07\ \text{V} $

Since $+0.07\ \text{V} > -0.88\ \text{V}$, the $\mathrm{AgBr/Ag}$ half-cell will undergo reduction and the iron half-cell will undergo oxidation.

So reverse the iron half-reaction:

$ \mathrm{Fe(s) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) + 2e^-} $

Multiply the silver half-reaction by $2$:

$ \mathrm{2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq)} $

Now add them:

$ \mathrm{Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)} $

This matches Option A.

Now calculate $E^\theta_{\text{cell}}$:

$ E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} $

Here,

• cathode: $\mathrm{AgBr/Ag}$, so $E^\theta_{\text{cathode}} = +0.07\ \text{V}$

• anode: reduction potential of $\mathrm{Fe(OH)_2/Fe}$ is $-0.88\ \text{V}$

Therefore,

$ E^\theta_{\text{cell}} = 0.07 - (-0.88) = 0.95\ \text{V} $

So Option B is false.

Option C is false because Fe is oxidised, not reduced.

Option D is false because $E^\theta_{\text{cell}}$ is an intensive property, not an extensive property.

Hence, the correct option is:

$ \boxed{\text{Option A}} $

Let’s analyze the situation step by step.

We have 1 M aqueous solutions of:

$\text{AgNO}_3$ with $\text{Ag}^+/ \text{Ag} \quad E^0 = +0.80\,V$

$\text{Hg}_2(\text{NO}_3)_2$ with $\text{Hg}_2^{2+}/ \text{Hg} \quad E^0 = +0.79\,V$

$\text{Cu(NO}_3)_2$ with $\text{Cu}^{2+}/ \text{Cu} \quad E^0 = +0.24\,V$

$\text{Mg(NO}_3)_2$ with $\text{Mg}^{2+}/ \text{Mg} \quad E^0 = -2.37\,V$

In an electrolytic cell with inert electrodes, reduction occurs at the cathode. The ion with the highest (most positive) reduction potential will be reduced (or deposited) first. Therefore, as we increase the applied voltage (making the cathode sufficiently negative), the metals will begin to deposit in the order of their reduction potentials.

The order of reduction (deposition) will be:

First, silver ($+0.80\,V$)

Next, mercury ($+0.79\,V$)

Then, copper ($+0.24\,V$)

So, Statement (I) which says “With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu” is correct.

For magnesium, the reduction potential is very negative ($-2.37\,V$). In aqueous solution, before reaching such a negative potential, water will be reduced. The cathodic reaction for water typically is:

$2\,\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\,\text{OH}^-$

which leads to the evolution of hydrogen gas, not oxygen. (Oxygen is produced at the anode during the oxidation of water.)

Hence, Statement (II) erroneously states that “oxygen gas will be evolved at the cathode” when in fact, if magnesium were to be reduced (which it won’t be), water reduction would yield hydrogen gas. Thus, Statement (II) is incorrect.

Summary:

Statement (I) is correct.

Statement (II) is incorrect.

Therefore, the most appropriate answer is:

Option C: Statement I is correct but Statement II is incorrect.

For charging of lead storage battery cell reaction is

$2 \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Pb}(\mathrm{~s})+\mathrm{PbO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})$

At anode $\mathrm{PbSO}_4$ reduced back to Pb and at cathode $\mathrm{PbSO}_4$ oxidised back to $\mathrm{PbO}_2$.

$\begin{aligned} \because ~& \mathrm{x}_1=+2, \mathrm{y}_1=0 \\ \mathrm{x}_2 & =+2, \mathrm{y}_2=4 \end{aligned}$

To determine the reaction potentials at the anode and cathode in the methanol fuel cell, we start with the given standard cell potential:

$ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} $

Given:

$ E_{\text{cell}}^{\circ} = 1.21 \, \text{V} $

$ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} = 1.229 \, \text{V} $

Substituting the values,

$ 1.21 = 1.229 - E_{\text{anode}}^{\circ} $

Solving for $ E_{\text{anode}}^{\circ} $:

$ E_{\text{anode}}^{\circ} = 1.229 - 1.21 = 0.019 \, \text{V} $

Since this is a fuel cell based on the oxidation of methanol, methanol undergoes oxidation at the anode, while oxygen is reduced at the cathode. Therefore:

Anode Reaction: Oxidation of methanol occurs here, contributing to the fuel cell’s electrical output.

Cathode Reaction: Reduction of oxygen occurs at the cathode, where the $ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} $ potential is utilized.

Thus, we verify that the half-cell reduction potential for oxidation of methanol by CO$_2$ is indeed 19 mV, aligning with the calculated $ E_{\text{anode}}^{\circ} $. The processes correctly identify the roles of anode and cathode in this fuel cell.

The limiting molar conductivities of ions are important in determining how well they conduct electricity in solution. Here are the values for some common cations at 298 K:

H⁺: 349.8 S cm² mol⁻¹

Na⁺: 50.11 S cm² mol⁻¹

K⁺: 73.52 S cm² mol⁻¹

Ca²⁺: 119 S cm² mol⁻¹

Mg²⁺: 106.12 S cm² mol⁻¹

Based on these values, the correct order of limiting molar conductivity for these cations is:

$ \stackrel{\oplus}{\text{H}}^+ > \text{Ca}^{2+} > \text{Mg}^{2+} > \text{K}^+ > \text{Na}^+ $

This order reflects the decreasing ability of each ion to conduct electricity in water under the given conditions.

For each application listed in List-I, match it with the appropriate battery or cell type from List-II based on their characteristics:

(A) Transistors: Generally use a dry cell or Leclanché cell, where the anode is zinc (Zn) and the cathode is a carbon rod surrounded by manganese dioxide (MnO₂).

Therefore, (A) Transistors match with (III) Anode - Zn; Cathode - Carbon.

(B) Hearing aids: Operate using a mercury cell. The anode in this cell is a zinc-mercury amalgam, and the cathode is composed of mercury oxide (HgO) with carbon.

Thus, (B) Hearing aids match with (I) Anode - Zn/Hg; Cathode - HgO + C.

(C) Inverters: Use a lead-acid storage cell. In this type of cell, the anode consists of lead (Pb), and the cathode is a lead dioxide (PbO₂) grid.

As a result, (C) Inverters match with (IV) Anode - Pb; Cathode - Pb | PbO₂.

(D) Apollo space ship: Utilizes a hydrogen-oxygen fuel cell. The hydrogen is oxidized at the anode, while oxygen is reduced at the cathode.

Therefore, (D) Apollo space ship matches with (II) Hydrogen fuel cell.

Summary of matches:

A - III

B - I

C - IV

D - II

This corresponds with Answer Option A.

(A) An aqueous solution of AgNO$_3$ using silver electrodes.

Cathode - Reduction - $Ag_{(aq)}^ + + {e^ - } \to Ag(s)$

Anode - Oxidation - $Ag(s) \to Ag_{(aq)}^ + + {e^ - }$

Solid silver will be deposited at the cathode. Solid anode (silver) will dissolve, releasing silver ions into the solution.

So, there is no formation of O$_2$ gas in this electrolysis.

(B) An aqueous solution of AgNO$_3$ using platinum electrodes.

Cathode - Reduction - $Ag_{(aq)}^ + + {e^ - } \to Ag(s)$

Anode - Oxidation - $2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$

When platinum electrodes are used, Ag$^+$ from solution is reduced and deposited at cathode whereas O$_2$ is produced at the anode.

(C) A dilute soution of H$_2$SO$_4$ using platinum electrodes.

Cathode - Reduction - $2{H^ + } + 2{e^ - } \to {H_2}$

Anode - Oxidation - $2{H_2}O \to {O_2} + 4{H^ + } + 4{e^ - }$

H$_2$ gas is produded at the cathode and O$_2$ gas is produced at the anode.

(D) a high concentration solution of H$_2$SO$_4$ using platinum electrodes.

O$_2$ gas is not formed in this case.

Cathode - Reduction - The substance formed is H$_2$ gas.

Anode - Oxidation - The substance formed is not O$_2$ gas.

So, statements (B) and (C) are correct.

$\mathop {Mg\left| {M{g^{2 + }}(aq)} \right|}\limits_{(Anode)} \mathop {\left| {A{g^ + }(aq)} \right|Ag}\limits_{(Cathode)} $

Nernst equation (general)

$E_{Cell}=E_{Cell}^0-\frac{R T}{n F} \ln Q$

For the cell, the chemical equation can be written as

$\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\text {aq })}^{+} \longrightarrow \mathrm{Mg}_{(\text {aq })}^{2+}+2 \mathrm{Ag}(\mathrm{~s})$

$n$ (number of electrons transferred $)=2$

${E_{cell}} = E_{cell}^o - {{RT} \over {2F}}\ln {{\left[ {M{g^{2 + }}} \right]{{\left[ {Ag} \right]}^2}} \over {\left[ {Mg} \right]{{\left[ {A{g^ + }} \right]}^2}}}$

${E_{cell}} = E_{cell}^o - {{RT} \over {2F}}\ln {{\left[ {M{g^{2 + }}} \right]} \over {{{\left[ {A{g^ + }} \right]}^2}}}$

${E_{cell}} = E_{cell}^o + {{RT} \over {2F}}\ln {{{{\left[ {A{g^ + }} \right]}^2}} \over {\left[ {M{g^{2 + }}} \right]}}$

Concentration of solid species

$[\mathrm{Mg}]=[\mathrm{Ag}]=1$

Correct answer :

Option D) ${E_{cell}} = E_{cell}^o + {{RT} \over {2F}}\ln {{{{\left[ {A{g^ + }} \right]}^2}} \over {\left[ {M{g^{2 + }}} \right]}}$

The molar conductivity of a weak electrolyte is plotted against the square root of its concentration:

$c$-concentration of weak electrolyte

Weak electrolytes give curved decreasing graph.

1) Molar conductivity increases sharply with increase in concentration.

Incorrect.

With increase in concentration, molar conductivity decreases and it is lean from the graph of molar conductivity vs $\sqrt{c}$.

2) Molar conductivity decreases sharply with increase in concentration.

Correct

From the graph it is clean that, molar conductivity decreases sharply with increase in concentration.

3) A small increase in molar conductivity is observed at infinite dilution.

Incorrect

There is no small increase in molas conductivity with increase in concentration for a weak electrolyte. And there is no small increase in mola conductivity. Significant increase occurs.

4) A small decrease in molar conductivity is observed at infinite dilution.

Incorrect

For a weak electrolyte, the molar conductivity significantly increases with dilution, especially at infinite dilution.

Correct answer:

Option 2) Molen conductivity decreases sharply with increase in concentration.

Oxidizing capacity is an element's tendency to donate electrons, or the atmosphere's ability to oxidize compounds.

A higher standard reduction potential indicates a stronger oxidizing capacity. A substance with a higher standard reduction potential is move likely to gain electrons and act as an oxidizing agent in a redon reaction. The more positive the reduction potential, the stronger the oxidizing power of a species.

From the given options, the higher standard reduction potential is for lead $(\mathrm{Pb}) .(+1.67)$.

so, the p-block ion with strongest oxidizing capacity is Pb (lead).

Answer: Option 3) $E_{p_b{ }^{4+} / \mathrm{Pb}^{2+}}^0=+1.67 \mathrm{~V}$

To determine the strongest reducing agent among the options, we need to consider which species most readily donates electrons (i.e., is most easily oxidized). One standard method is to compare the standard reduction potentials of the corresponding half-reactions. A lower (more negative) standard reduction potential indicates that the reduced species is less stable and more prone to oxidation.

Here’s the data provided:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7H_2O,\quad E^\circ = +1.33\,V$

$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V$

$\mathrm{MnO}_4^- + 8H^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4H_2O,\quad E^\circ = +1.51\,V$

$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V$

Now, let’s consider the species from the Options:

Option A: Cl⁻

The relevant half-reaction is:

$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V.$

The reduced form here is Cl⁻. Its corresponding oxidation (the reverse reaction) would have an electrode potential of

$E^\circ_{\text{oxidation}} = -1.36\,V.$

Option B: MnO₄⁻

In the given half-reaction, MnO₄⁻ acts as the oxidizing agent (it is reduced to Mn²⁺). Therefore, MnO₄⁻ is not a good reducing agent.

Option C: Cr (metallic chromium)

The relevant half-reaction is:

$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V.$

Here, Cr is the reduced species. Reversing this reaction gives:

$\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^-,$

with an effective oxidation potential of

$E^\circ_{\text{oxidation}} = +0.74\,V.$

Option D: Mn²⁺

In the MnO₄⁻/Mn²⁺ couple, Mn²⁺ is the reduced form. Its reverse (oxidation) potential would be:

$E^\circ_{\text{oxidation}} = -1.51\,V.$

When comparing the oxidation potentials (remember, a higher oxidation potential means the species is more willing to lose electrons), we have:

Cl⁻: $E^\circ_{\text{ox}} = -1.36\,V$

Cr: $E^\circ_{\text{ox}} = +0.74\,V$

Mn²⁺: $E^\circ_{\text{ox}} = -1.51\,V$

Among these, metallic Cr stands out because its corresponding reduction half-reaction has the most negative potential ($-0.74\,V$). This negative value indicates that Cr (in its elemental form) is unstable relative to its oxidized form ($\mathrm{Cr}^{3+}$) and, therefore, is the most eager to lose electrons. In other words, Cr is the strongest reducing agent.

Thus, the answer is:

Option C: $\mathrm{Cr}$

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

$\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}$

Also $\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0$

$\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}$

$\mathrm{E}_{\text {Cell }}^0$ for reaction will be

$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}^0}^0+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0 \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}$

Option (2)

$\begin{gathered} \because \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ \text { Option }(1) \mathrm{E}^{\circ}=0.8+0.76 \\ =1.56 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 1.56 \\ =-3.12 \mathrm{~V} \\ \text { Option (2) } \mathrm{E}^{\circ}=-2.37+0.76 \\ =-1.61 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times(-1.61) \\ =+3.22 \mathrm{~V} \\ \text { Option (3) } \mathrm{E}^{\circ}=-2.37-0.8 \\ =-3.17 \mathrm{~V} \\ \begin{array}{c} \therefore \Delta \mathrm{G}^{\circ}=-2 \end{array} \\ =\mathrm{F} \times(-3.17) \\ =+6.34 \\ \text { Option }(4) \mathrm{E}^{\circ}=0.8-0.34 \\ =0.46 \mathrm{~V} \\ \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 0.46 \\ =-0.92 \mathrm{~V} \end{gathered}$

$\begin{aligned} & \Delta \mathrm{G}_4^{\mathrm{o}}=\Delta \mathrm{G}_1^{\mathrm{o}}+\Delta \mathrm{G}_2^{\mathrm{o}} \\ \Rightarrow & -\mathrm{n}_4 \mathrm{FE}_4^{\mathrm{o}}=-\mathrm{n}_1 \mathrm{FE}_1^0-\mathrm{n}_2 \mathrm{FE}_2^{\mathrm{o}} \\ \Rightarrow & +4 \mathrm{E}_4^{\mathrm{o}}=3 \times 2+(1 \times 0.8) \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=\frac{6.8}{4} \mathrm{~V} \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=1.7 \mathrm{~V} \end{aligned}$

Let’s examine both statements one by one and see whether they hold true for typical corrosion processes (e.g., iron rusting):

Statement (I)

"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode."

Corrosion is electrochemical

True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and reduction at another region (the cathode) on the metal’s surface.

Pure metal as anode, impurity as cathode

In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) — that is, it serves as the anode.

The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode region where reduction (e.g., oxygen reduction) occurs.

This difference in electrode potentials between the pure region and the impurity region drives the corrosion cell.

Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of iron).

Statement (II)

"The rate of corrosion is more in alkaline medium than in acidic medium."

For most common metals (like iron), acidic media typically enhance corrosion because abundant H⁺ ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal.

In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide layers that can slow down further corrosion.

While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron and many other metals, corrosion is more severe in acidic environments than in alkaline ones.

Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron).

Conclusion

Statement (I): True

Statement (II): False

Therefore, the correct choice (matching these truth values) is:

(C) Statement I is true but Statement II is false.

$\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ is obtained by electrolysis of concentrated solution of sulphuric acid.

At anode :

$ 2 \mathrm{HSO}_4^{-} \rightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{e}^{-} $

$ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} $

The mass of aluminium deposited can be calculated using Faraday’s laws of electrolysis. The steps are as follows:

Compute the total charge passed:

$ Q = I \cdot t = 2\,\text{A} \cdot (30 \times 60\,\text{s}) = 3600\,\text{C} $

Determine the moles of aluminium deposited. Since three electrons are required to deposit one mole of aluminium, the number of moles is given by:

$ n(\text{Al}) = \frac{Q}{3F} = \frac{3600}{3 \times 96500} \approx 0.01242\,\text{mol} $

Finally, calculate the mass of aluminium using its molar mass ($M = 27\,\text{g/mol}$):

$ m(\text{Al}) = n(\text{Al}) \times M = 0.01242 \times 27 \approx 0.335\,\text{g} $

Thus, the amount of aluminium deposited at the cathode is approximately

$ 0.336\, \text{g} $

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} $

Using the Nernst equation,

$ E=E^{\circ}-\frac{2303}{n F} \log Q $

For the reduction,

$ \begin{aligned} & \mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \\ & a=\frac{t}{\left[\mathrm{Cu}^{2+}\right]} \\ & E=E^{\circ}+\frac{2303}{n F} \log \left[\mathrm{Cu}^{2+}\right] \end{aligned} $

Substituting the values, $n=2$

$ \begin{aligned} & E^{\circ}=0.34 \mathrm{~V},\left[\mathrm{Cu}^{2+}\right]=0.1 \\ & E=0.34+\frac{0.056}{2} \log (0.1) \\ & E=0.31 \mathrm{~V} \end{aligned} $

Among given statements regardiner dry cell, I, III and IV are correct. While II is incorrect. It's correct form is, in dry cell, zinc vessel act as anode.

Let's match each symbol of the electrical property with its corresponding unit:

(A) $\Lambda_m$ (Molar conductivity)

Molar conductivity ($\Lambda_m$) is defined as the conductivity ($\kappa$) divided by the molar concentration ($c$).

$\Lambda_m = \frac{\kappa}{c}$

Units of $\kappa$ are S cm⁻¹.

Units of $c$ are mol cm⁻³.

Therefore, the units of $\Lambda_m$ are $\frac{\text{S cm}^{-1}}{\text{mol cm}^{-3}} = \text{S cm}^2 \text{ mol}^{-1}$.

So, (A) matches with (III).

(B) G (Conductance)

Conductance (G) is the reciprocal of resistance (R).

$G = \frac{1}{R}$

The unit of resistance is Ohm ($\Omega$).

Therefore, the unit of conductance is Ohm⁻¹ or Siemens (S).

So, (B) matches with (IV).

(C) $\kappa$ (Kappa, Conductivity or Specific conductance)

Conductivity ($\kappa$) is the reciprocal of resistivity ($\rho$). It is the conductance (G) of a material with unit length ($l=1$) and unit cross-sectional area ($A=1$).

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

Units of G are S.

Units of $l$ are cm.

Units of $A$ are cm².

Therefore, the units of $\kappa$ are $\text{S} \times \frac{\text{cm}}{\text{cm}^2} = \text{S cm}^{-1}$.

So, (C) matches with (I).

(D) $G^*$ (Cell constant)

The cell constant ($G^*$) for a conductivity cell is defined as the ratio of the distance between the electrodes ($l$) to the area of the electrodes ($A$).

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

Units of $l$ are cm.

Units of $A$ are cm².

Therefore, the units of $G^*$ are $\frac{\text{cm}}{\text{cm}^2} = \text{cm}^{-1}$.

Alternatively, we know that conductivity $\kappa = G \times G^*$, so $G^* = \frac{\kappa}{G}$.

Units of $\kappa$ are S cm⁻¹.

Units of G are S.

So, units of $G^*$ are $\frac{\text{S cm}^{-1}}{\text{S}} = \text{cm}^{-1}$.

So, (D) matches with (II).

Summarizing the matches:

(A) - (III)

(B) - (IV)

(C) - (I)

(D) - (II)

The reaction is,

$ \begin{aligned} & 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{I}_2(s) \\ & n=2 \end{aligned} $

At equilibrium, $E_{\text {cell }}=0 \mathrm{~V}, Q=K$

Using Nernst equation

$ E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln K $

$\ln K=\frac{n F E_{\text {cell }}^{\circ}}{R T}$, substituting values, $\ln$

$ \begin{aligned} K & =18.493 \\ \log _{10} K & =\frac{18.493}{2.303} \approx 8 \Rightarrow 10^8 \end{aligned} $

So, the value of $x$ is 8 .

During discharge of a lead storage battery, the reaction at the anode involves the oxidation of lead into lead sulphate $\left(\mathrm{PbSO}_4\right)$ and releases electron.

$ \mathrm{Pb}(s)+\mathrm{SO}_4^{2-}(a q) \longrightarrow \mathrm{PbSO}_4(s)+2 e^{-} $

Let's match the standard electrode potentials ($E^\ominus_{M^{2+}/M}$) for the given transition metals. We need to recall the typical values for these elements from the electrochemical series.

The standard reduction potentials for the given metals are approximately:

• Manganese (Mn): $E^\ominus_{Mn^{2+}/Mn}$ is known to be significantly negative, typically -1.18 V. This is due to its strong tendency to form Mn²⁺, especially considering the stable half-filled d-subshell (d⁵) in Mn²⁺ after losing two electrons from its [Ar]3d⁵4s² configuration.

• Chromium (Cr): $E^\ominus_{Cr^{2+}/Cr}$ is also quite negative, typically -0.91 V.

• Iron (Fe): $E^\ominus_{Fe^{2+}/Fe}$ is moderately negative, typically -0.44 V.

• Nickel (Ni): $E^\ominus_{Ni^{2+}/Ni}$ is the least negative among the given options, typically -0.25 V.

Now let's match these values with the given List-II:

(A) Ni: Matches with (III) -0.25 V

(B) Mn: Matches with (I) -1.18 V

(C) Fe: Matches with (IV) -0.44 V

(D) Cr: Matches with (II) -0.91 V

So, the correct matching is:

A - III

B - I

C - IV

D - II

Let's check the given options:

• Option A: A-III, B-I, C-IV, D-I (Incorrect, D is not I)

• Option B: A-III, B-IV, C-I, D-II (Incorrect, B is not IV, C is not I)

• Option C: A-III, B-I, C-IV, D-II (This matches our derived mapping)

• Option D: A-I, B-IV, C-II, D-III (Incorrect, A is not I, B is not IV, C is not II, D is not III)

Therefore, Option C is the correct answer.

The final answer is $\boxed{\text{Option C}}$

Step 1: Write the reaction and information given

The reaction is: $ \mathrm{H}^{+}(aq) + e^{-} \rightarrow \frac{1}{2} \mathrm{H}_2 \ (1\ \text{bar}) $ The pH of the solution is 10.0.

Step 2: Find the concentration of $ \mathrm{H}^{+} $

We know that $\mathrm{pH} = -\log [\mathrm{H}^{+}]$

This means $[\mathrm{H}^{+}] = 10^{-\mathrm{pH}} = 10^{-10}\ \mathrm{M}$

Step 3: Note important values

Standard electrode potential for the hydrogen electrode is $E^\circ = 0$ V and the number of electrons ($n$) is 1.

Step 4: Write and use the Nernst Equation

The Nernst equation is: $ E = E^\circ - \frac{0.0591}{n} \log [\mathrm{H}^{+}] $ Now plug in the values: $E = 0 - \frac{0.0591}{1} \log (10^{-10})$

Step 5: Calculate the answer

$\log (10^{-10}) = -10$, so $E = 0 - (0.0591 \times -10)$

$E = 0 + 0.591\ \mathrm{V}$

Step 6: Final value and sign

The calculated value is $E = 0.591~\mathrm{V}$. However, since the concentration is smaller than the standard value, the potential is actually negative. So, $E = -0.591~\mathrm{V}$ or approximately $-0.6~\mathrm{V}$.

Step 1: Find the cell constant

The cell constant tells us about the cell's design and size. It can be found by using the known conductivity and resistance of 0.1 M KCl solution.

$ G^{\star} = \kappa \cdot R $

So, $ G^{\star} = 1.29 \times 100 = 129\ \text{m}^{-1} $ or $ 1.29\ \text{cm}^{-1} $

Step 2: Calculate the conductivity ($\kappa$) for 0.02 M KCl solution

We use the cell constant and the new measured resistance for the 0.02 M solution.

$ \kappa = \frac{G^{\star}}{R} = \frac{1.29}{520} = 0.00248\ \mathrm{Sm}^{-1} $

Step 3: Find the molar conductivity ($\Lambda_m$)

Molar conductivity shows how well ions move in a solution. Use this formula:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

$ \begin{aligned} \Lambda_m &= \frac{0.00248 \times 1000}{0.02} \\ &= 124~\mathrm{Scm}^2~\mathrm{mol}^{-1} \end{aligned} $

Among the given options, Cr has $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}$ negative.

The value of $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$.

readily oxidised to $\mathrm{Cr}^{3+}$ with a relatively negative potential.

Oxidation reaction

$ \mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 e $

Reduction $2 \mathrm{Fe}^{3+}+2 e^{-} \longrightarrow 2 \mathrm{Fe}^{2+}$

$ \begin{aligned} E_{\text {anode }} & =E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.441 \mathrm{~V}, \\ E_{\text {cathode }} & =0.771 \mathrm{~V} \\ E_{\text {cell }} & =E_{\text {cathode }}-E_{\text {anode }} \\ & =0.771 \mathrm{~V}-(-0.441 \mathrm{~V}) \\ & =1.212 \mathrm{~V} \end{aligned} $

Given, $\kappa=0.0115 \mathrm{Sm}^{-1}$

Concentration of NaOH solution $=0.05 \mathrm{M}$ or molar conductance, $\Lambda_m$ is given by,

$ \begin{aligned} \Lambda_m & =\frac{\kappa}{C} \times 1000 \\ & =\frac{0.0115}{0.050} \times 1000 \mathrm{in} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \Lambda_m & =230 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned} $

Using Nernst equation,

$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Ni}^{2+}}\right] \\ & \begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {Cathode }}^{\circ}-E_{\text {Anode }}^{\circ} \\ & =E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} \\ & =(-0.25 \mathrm{~V})-(-0.76 \mathrm{~V})=0.51 \mathrm{~V} \end{aligned} \end{aligned} $

Substituting all the given values

$ \begin{aligned} E_{\text {cell }} & =0.51-\frac{0.06}{2} \log \left(\frac{0.01}{0.1}\right) \\ & =0.51-0.03 \log (0.1) \\ & =0.51+0.03=0.54 \mathrm{~V} \end{aligned} $

Electrolysis of NaCl produces NaOH at the cathode.

Charge passed $=4 \times 1200=4800 \mathrm{C}$

Moles of electrons $=\frac{4800}{96500} \approx 0.04974 \mathrm{~mol}$

Since 2 moles of electron produce 2 moles NaOH , NaOH formed $\boldsymbol{\approx} 0.04974 \mathrm{~mol}$

$ \begin{gathered} \text { Concentration }=\frac{0.04974}{0.5}=0.0995 \mathrm{M} \\ \text { pOH }=-\log 0.0995 \\ \text { pOH } \approx 1, \text { so } \mathrm{pH} \approx 14-1=13 \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \mathrm{Ca}^{2+}+2 e^{-} \quad \longrightarrow \mathrm{Ca} \text { (at cathode) } \\ & 1 \text { mole of } \mathrm{CaCl}_2=2 \mathrm{~F} \\ & \,\,\,\,\,\,\, \quad Q=I t \Rightarrow n \cdot 2 F=I t \\ & \Rightarrow \quad t=\frac{n \cdot 2 F}{I}=\frac{20}{40} \times \frac{2 \times 96500}{4} \\ &  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=24125 \mathrm{~s}=6.7 \text { hours. } \end{aligned} \end{aligned} $

To find the correct match between List I (Cell) and List II (Use/Property/Reaction), let's first understand each item and its corresponding match:

List I (Cell):

• Leclanché cell: This is a primary cell (non-rechargeable) widely used in flashlights and other portable devices. It's not designed for recharging.


• Ni - Cd cell: Nickel-Cadmium (NiCd) batteries are rechargeable and known for being used in portable electronics, tools, and various battery-operated devices.


• Fuel cell: Converts chemical energy, from a fuel into electricity through a chemical reaction with oxygen or another oxidizing agent. It converts energy of combustion (such as hydrogen) into electrical energy directly.


• Mercury cell: A type of battery that uses mercury oxide. It's known for its stable output voltage, making it suitable for devices like hearing aids. However, the reaction described in the list II does not pertain to the mercury cell but more closely to the Leclanché cell or similar.

List II (Use/Property/Reaction):

• Converts energy of combustion into electrical energy: This directly relates to the principle behind fuel cells, which generate electricity through the chemical reaction of a fuel with oxygen.


• Does not involve any ion in solution and is used in hearing aids: This refers to the mercury cell, which is chosen for devices requiring a stable voltage, like hearing aids, although the description about ions might be misleading. The characteristic here is about stable voltage and application in hearing aids.


• Rechargeable: This trait is indicative of Ni - Cd cells, which are known for their ability to be recharged multiple times.


• Reaction at anode $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$: This anodic reaction is typical in cells like the Leclanché cell, which features a zinc anode undergoing oxidation.

Given these explanations, the correct matches would be:

• A (Leclanché cell) with IV, due to the anodic reaction involving zinc.


• B (Ni - Cd cell) with III, because it is rechargeable.


• C (Fuel cell) with I, as it converts the energy of combustion into electrical energy.


• D (Mercury cell) with II, by elimination and considering its use in devices like hearing aids and the slight misdescription regarding ions.

Therefore, the correct answer is Option C: A-IV, B-III, C-I, D-II.