Electrochemistry

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} $

Given the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

The standard cell potential is:

$ E^{\ominus} = 1 \, \text{V} $

The standard entropy change is:

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $

The temperature is:

$ T = 300 \, \text{K} $

There is a relationship between the change in Gibbs free energy, entropy, and enthalpy, given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Also, the change in Gibbs free energy is related to the cell potential and the number of moles of electrons transferred ($n$) through the equation:

$ \Delta_r G^{\ominus} = -n F E^{\ominus} $

Where $ F $ is the Faraday constant ($96500 \, \text{C/mol}$).

Substituting into the equations, we find:

$ -n F E^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Rearranging and substituting the known values:

$ -2 \times 96500 \times 1 = \Delta_r H^{\ominus} - 300 \times 100 $

Solving this, we get:

$ \Delta_r H^{\ominus} = -163 \times 10^3 \, \text{J/mol} $

Therefore:

$ \Delta_r H^{\ominus} = -163 \, \text{kJ/mol} $

To find the change in entropy ($\Delta_r S^{\ominus}$) for the given cell reaction at 300 K, we start with the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

Given parameters are:

Standard cell potential, $E^{\ominus} = 1.0 \, \text{V}$

Standard reaction enthalpy change, $\Delta_r H^{\ominus} = -163 \, \text{kJ/mol}$

Temperature, $T = 300 \, \text{K}$

Faraday's constant, $F = 96500 \, \text{C/mol}$

The relationship between Gibbs free energy change ($\Delta_r G^{\ominus}$), enthalpy change ($\Delta_r H^{\ominus}$), and entropy change ($\Delta_r S^{\ominus}$) is given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Additionally, the change in Gibbs free energy for the cell is also related to the cell potential by the equation:

$ \Delta_r G^{\ominus} = -nFE^{\ominus} $

where $n$ is the number of moles of electrons exchanged in the reaction. Here, $n = 2$ since two electrons are transferred.

Substituting the known values into the equation:

$ -nFE^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Calculate as follows:

$ -2 \times 96500 \times 1 = -163000 - T \Delta_r S^{\ominus} $

Simplifying:

$ -193000 = -163000 - 300 \Delta_r S^{\ominus} $

Rearranging to solve for $\Delta_r S^{\ominus}$:

$ 30000 = 300 \Delta_r S^{\ominus} $

$ \Delta_r S^{\ominus} = \frac{30000}{300} $

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $

$ \text { The cell reaction is, } $

$ \begin{aligned} & \therefore \quad Q=\frac{\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2}{\mathrm{p}_{\mathrm{Cl}_2}} \\ & \text { and } E_{\text {cell }}=E^{\circ}-\frac{0.0591}{2} \log Q \\ & \text { So, } E_{\text {cell }} \propto \frac{1}{Q} \end{aligned} $

(a) $Q=\frac{1 \times 1^2}{1}=1$;

(b) $Q=\frac{0.01 \times 1^2}{1}=10^{-2}$

(c) $Q=\frac{1 \times(0.01)^2}{1}=10^{-4}$;

(d) $Q=\frac{0.01 \times(0.01)^2}{1}=10^{-6}$

$\Rightarrow$ The order of the values of

$ Q: d

$\Rightarrow$ The order of emf of galvanic cells:

$ d>c>b>a $

The emf is maximum in

$ \begin{aligned} & \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(0.01 \mathrm{M})\right| \\ & \mathrm{Cl}_2(1 \mathrm{~atm}) \mid \mathrm{Pt} \end{aligned} $

Let's analyze the data step by step.

We are given the following standard electrode reactions and their potentials:

For aluminum (Al):

$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}, \quad E^\circ = -1.66 \text{ V}$

$\mathrm{Al}^{+} + e^- \to \mathrm{Al}, \quad E^\circ = +0.55 \text{ V}$

For thallium (Tl):

$\mathrm{Tl}^{3+} + 3e^- \to \mathrm{Tl}, \quad E^\circ = +1.26 \text{ V}$

$\mathrm{Tl}^{+} + e^- \to \mathrm{Tl}, \quad E^\circ = -0.34 \text{ V}$

A key idea in interpreting these potentials is that a more positive reduction potential means the ion is a stronger oxidizing agent and is more eager to gain electrons (i.e., it is less “stable” in its oxidized form). In contrast, a more negative reduction potential indicates that the species is less likely to gain electrons and is more stable in that oxidation state.

Let’s apply this reasoning:

• For aluminum:

The reaction $\mathrm{Al}^{3+} \to \mathrm{Al}$ has a very negative potential (–1.66 V), meaning that once aluminum is in the +3 state, it does not easily pick up electrons to revert to the metal. This makes $\mathrm{Al}^{3+}$ a very stable oxidation state.

The reaction involving $\mathrm{Al}^{+}$ has a positive potential (+0.55 V), indicating that $\mathrm{Al}^{+}$ is readily reduced to aluminum, and thus is less stable in the +1 state.

• For thallium:

The reaction $\mathrm{Tl}^{3+} \to \mathrm{Tl}$ has a highly positive potential (+1.26 V), so $\mathrm{Tl}^{3+}$ is not very stable; it tends strongly to gain electrons.

In contrast, the reaction involving $\mathrm{Tl}^{+}$ has a negative potential (–0.34 V), which indicates that $\mathrm{Tl}^{+}$ does not easily accept an electron, making it a relatively stable oxidation state.

Now, let’s look at the options:

Option A: “$\mathrm{Tl}^{3+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Because $\mathrm{Tl}^{3+}$ (with +1.26 V) is eager to gain electrons (a strong oxidizing agent), it is less stable compared to $\mathrm{Al}^{3+}$ (with –1.66 V). Thus, Option A is incorrect.

Option B and Option C: “$\mathrm{Al}^{+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Our data indicates that $\mathrm{Al}^{+}$ (with +0.55 V) is less stable (more readily reduced) than $\mathrm{Al}^{3+}$ (with –1.66 V). So, these statements are not correct.

Option D: “$\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.”

• Comparing the two, the reduction reaction for $\mathrm{Tl}^{+}$ has $E^\circ = -0.34\text{ V}$ while that for $\mathrm{Al}^{+}$ has $E^\circ = +0.55\text{ V}.$ The more negative potential for $\mathrm{Tl}^{+}$ shows it is less likely to be reduced to elemental Tl, meaning $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Thus, Option D is the correct statement.

In summary, the correct answer is:

Option D: $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Let's analyze each statement using the given standard reduction potentials:

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

$Cu^{2+} + 2e^- \rightarrow Cu,\quad E^\circ = +0.34\text{ V}$

$Zn^{2+} + 2e^- \rightarrow Zn,\quad E^\circ = -0.76\text{ V}$

For nitrate reducing to nitric oxide (in acidic solution):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Now, let’s consider each statement one by one.

Statement I: “$H^+$ does not oxidise Cu to $Cu^{2+}$”

The possible reaction would be:

$Cu(s) + 2H^+ \rightarrow Cu^{2+} + H_2(g)$

To check spontaneity, write the half-reactions:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,$

(reverse of the given reduction reaction with $E^\circ = -0.34\text{ V}$)

Reduction (hydrogen):

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

The cell potential is:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.0\text{ V} - 0.34\text{ V} = -0.34\text{ V}$

A negative cell potential means the reaction is non-spontaneous.

Thus, Statement I is correct.

Statement II: “Zn reduces $Cu^{2+}$ to Cu”

The reaction is:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Write the half-reactions:

Oxidation (zinc):

$Zn(s) \rightarrow Zn^{2+} + 2e^-,$

The oxidation potential for Zn is the reverse of its reduction: $+0.76\text{ V}$.

Reduction (copper):

$Cu^{2+} + 2e^- \rightarrow Cu(s),\quad E^\circ = +0.34\text{ V}$

Combine the potentials:

$E^\circ_{\text{cell}} = 0.34\text{ V} + 0.76\text{ V} = +1.10\text{ V}$

A positive cell potential indicates the reaction is spontaneous.

Thus, Statement II is correct.

Statement III: “$NO_3^-$ oxidises Cu to $Cu^{2+}$”

In reactions with nitric acid (which provides $NO_3^-$ in acidic medium), copper is oxidized. The half-reactions are:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,\quad E^\circ_{\text{ox}} = -0.34\text{ V}$

Reduction (nitrate in acid):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Even though the electrons do not match directly (2 vs. 3 electrons), the overall potential difference can be determined by the difference in the standard electrode potentials:

$E^\circ_{\text{net}} = 0.97\text{ V} - 0.34\text{ V} = +0.63\text{ V}$

A positive net potential shows that the oxidation of Cu by $NO_3^-$ in acidic conditions is thermodynamically favored.

Thus, Statement III is correct.

Since all three statements (I, II, and III) are correct, the correct option is:

Option B: I, II, III

A reducing agent is a substance that donates electrons during a chemical reaction. The more negative the standard reduction potential (given by the half-reaction), the stronger the tendency of the substance to lose electrons—making it a stronger reducing agent.

Let’s compare the given elements by their half-reactions and approximate standard reduction potentials:

For sodium (Na):

$\mathrm{Na^+ + e^- \to Na} \quad E^\circ \approx -2.71\,V$

For rubidium (Rb):

$\mathrm{Rb^+ + e^- \to Rb} \quad E^\circ \approx -2.98\,V$

For magnesium (Mg):

$\mathrm{Mg^{2+} + 2e^- \to Mg} \quad E^\circ \approx -2.36\,V$

For strontium (Sr):

$\mathrm{Sr^{2+} + 2e^- \to Sr} \quad E^\circ \approx -2.89\,V$

Since the strength of a reducing agent corresponds to having a very negative standard reduction potential, we compare the values:

Rb: approximately -2.98 V

Sr: approximately -2.89 V

Na: approximately -2.71 V

Mg: approximately -2.36 V

Rubidium (Rb) has the most negative reduction potential. Therefore, it is the strongest reducing agent among the given elements.

The correct answer is: Option A (Rb).

The standard reduction potentials for the following half-reactions are:

$2 \mathrm{H}^{+} / \mathrm{H}_2$: 0.0 V

$\mathrm{Cu}^{2+} / \mathrm{Cu}$: +0.34 V

$\mathrm{Zn}^{2+} / \mathrm{Zn}$: -0.76 V

$\mathrm{NO}_3^{-}, \mathrm{H}^{-} / \mathrm{NO}$: 0.97 V

Let's consider the reactions:

$\mathrm{Zn} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HNO}_3 \rightarrow$

We are tasked with identifying which of these reactions do not liberate $\mathrm{H}_2(g)$.

Reactions II and III do not produce $\mathrm{H}_2$ because they are not spontaneous. This conclusion is based on their cell potentials ($E_{\text{cell}}^{\circ}$) being negative ($E_{\text{cell}}^{\circ} < 0$).

Calculation of Cell Potentials:

Reaction II: $\mathrm{Cu} + \mathrm{HCl}$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.34 \\ &= -0.34 \, \text{V} \end{align*} $

Reaction III: $\mathrm{Cu} + \mathrm{HNO}_3$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.97 \\ &= -0.97 \, \text{V} \end{align*} $

Since both reactions have negative cell potentials, they are non-spontaneous and thus do not release $\mathrm{H}_2(g)$.

To determine the mass of copper deposited at the cathode during electrolysis, we first need to gather the given data and use the formula for electrolysis. The relevant parameters are:

Current, $ I = 2 \, \text{A} $

Time, $ t = 10 \, \text{minutes} $

Convert the time from minutes to seconds for consistency with the formula:

$ t = 10 \times 60 = 600 \, \text{seconds} $

Next, calculate the total charge $ Q $ passed during electrolysis using the formula:

$ Q = I \times t = 2 \times 600 = 1200 \, \text{C} $

The electrochemical reaction for copper deposition is:

$ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) $

Using the formula for the mass of substance deposited during electrolysis:

$ \text{Mass of copper deposited} = \frac{\text{Molar mass} \times \text{Charge}}{\text{Number of electrons transferred} \times F} $

Given:

Molar mass of Cu = 63 g/mol

Faraday constant, $ F = 96500 \, \text{C/mol} $

Number of electrons transferred for copper, = 2

Substitute these values into the formula:

$ \text{Mass of copper deposited} = \frac{63 \times 1200}{2 \times 96500} = 0.39 \, \text{g} $

Thus, the weight of copper deposited at the cathode is 0.39 grams.

To find the atomic mass of the metal $ M $, given that 2.644 g of metal was deposited with a total charge of 8040 Coulombs, we follow these steps:

Calculate the moles of electrons used:

$ \text{Moles of electrons} = \frac{\text{Total charge}}{\text{Faraday's constant}} = \frac{8040 \, \text{C}}{96500 \, \text{C/mol}} = 0.0833 \, \text{mol} $

Determine the moles of metal deposited:

The reaction $ M\mathrm{F}_2 \rightarrow M^{2+} + 2\mathrm{F}^- $ implies that 2 moles of electrons are needed to deposit 1 mole of metal.

$ \text{Moles of metal} (M) = \frac{\text{Moles of electrons}}{2} = \frac{0.0833}{2} = 0.04165 \, \text{mol} $

Calculate the atomic mass of metal $ M $:

$ \text{Atomic mass of } M = \frac{\text{Mass of metal}}{\text{Number of moles of metal}} = \frac{2.644 \, \text{g}}{0.04165 \, \text{mol}} \approx 63.47 $

Thus, the atomic mass of metal $ M $ is approximately 63.47 u.

In the electrolytic refining process of copper:

The anode is made up of impure copper. During the process, oxidation occurs at the anode, meaning the impure copper dissolves into the electrolyte.

The cathode is made of pure copper. Here, reduction takes place; copper ions gain electrons and deposit as pure copper on the cathode.

Mathematically, the reactions are as follows:

At the anode (oxidation):

$\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$

At the cathode (reduction):

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

Therefore, the correct option is:

Option B: impure copper, pure copper.

The given standard electrode potential is:

$ E^{\circ}_\text{{cell}} = 0.3 \, \text{V} $

According to the Nernst equation:

$ E_\text{{cell}} = E^{\circ}_\text{{cell}} - \frac{2.303RT}{nF} \log \left[\frac{\text{[Cu}^{2+}]}{\text{[M}^{2+}]}\right] \tag{1} $

In this context:

$ n = 2 $ (since it involves the transfer of 2 electrons)

$ E_\text{{cell}} = 0 $ (when the cell voltage is zero)

$\text{[M}^{2+}] = 0.1 \, \text{M} $

$\frac{2.303RT}{F} = 0.06 $

Substituting these values into equation (1) gives:

$ 0.3 = -\frac{0.06}{2} \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Solving for the log term:

$ -\frac{0.6}{0.06} = \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Taking the antilog:

$ 10^{-10} = \frac{\text{[Cu}^{2+}]}{0.1} $

This implies:

$ \text{[Cu}^{2+}] = 10^{-10} \times 10^{-1} = 10^{-11} \, \text{mol/L} $

Therefore, the concentration of $\text{Cu}^{2+}$ required to make the cell potential zero is $\boxed{10^{-11} \, \text{mol/L}}$.

Given :

$\begin{array}{ll} I = 96.5 \, \mathrm{A} \\ t = 100 \, \mathrm{s} \\ m = ? \end{array}$

The atomic weight of aluminum ($\mathrm{Al}$) is 27 u.

First, calculate the total charge (Q):

$\begin{aligned} Q &= I \times t \\ &= 96.5 \times 100 \\ &= 9650 \, \mathrm{C} \end{aligned}$

Consider the dissociation reaction of $\mathrm{AlCl}_3$:

$\begin{aligned} \mathrm{AlCl}_3 &\rightarrow \mathrm{Al}^{3+} + 3\mathrm{Cl}^{-} \\ \mathrm{Al}^{3+} + 3e^{-} &\rightarrow \mathrm{Al} \end{aligned}$

Understanding that 96,500 C (or 3 Faradays) of electricity deposits 1 mole of aluminum:

$\begin{aligned} \frac{9650 \, \mathrm{C}}{96,500 \, \mathrm{C/mol}} &= 0.0333 \, \mathrm{mol} \end{aligned}$

The weight of aluminum deposited can be calculated as:

$\begin{aligned} \text{Weight of } \mathrm{Al} &= n \times \text{atomic mass} \\ &= 0.0333 \times 27 \\ &\approx 0.90 \, \mathrm{g} \end{aligned}$

Given, current, $I=38.6 \mathrm{~A}$, Time, $t=100 \mathrm{~s}$

Cathode $\mathrm{Cu}^{2+}(a q)+2 e^\theta \longrightarrow \mathrm{Cu}(s)$

Anode $2 \overline{\mathrm{O}} \mathrm{H}(\mathrm{aq}) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_2(g)+2 e^{\ominus}$

Using Faraday's 1st law,

$W=Z I t, \quad(Z \rightarrow \text { electrochemical equivalent })$

$\begin{aligned} & =\frac{E I t}{96500}=\frac{M I t}{2 \times 96500} \\ & =\frac{63.5 \times 38.6 \times 100}{2 \times 96500}=1.27 \mathrm{~g} \end{aligned}$

1.27 g Cu is deposited.

Using Faraday's second law,

$\begin{aligned} & \frac{W_{\mathrm{Cu}}}{E_{\mathrm{Cu}}}=\frac{W_{\mathrm{O}_2}}{E_{\mathrm{O}_2}} \Rightarrow \frac{1.27}{63.5}=\frac{W_{\mathrm{O}_2}}{16} \\ & W_{\mathrm{O}_2}=\frac{1.27 \times 16}{63.5}=0.32 \mathrm{~g} \end{aligned}$

Moles of $\mathrm{O}_2=\frac{0.32}{32}=0.01 \mathrm{~mol}$

Volume of $\mathrm{O}_2$ at $\mathrm{STP}=0.01 \times 22.4=0.224 \mathrm{~L}$

$\begin{array}{r} \mathrm{H}^{+}(\mathrm{aq})+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(g) \\ p_{\mathrm{H}_2}=1 \mathrm{bar} \end{array}$

In a neutral electrode, concentration of $\mathrm{H}^{+}$ is $10^{-7} \mathrm{M}$.

Temperature, $T=298 \mathrm{~K}, E_{\mathrm{H}^{+} / \mathrm{H}_2 / \mathrm{Pt}}^{\circ}=0 \mathrm{~V}$

Using equation,

$\begin{aligned} & E=E_{\mathrm{H}^{+} / \mathrm{H}_2 / \mathrm{Pt}}^{\circ}-\frac{0.0591}{1} \log \frac{\left[p_{\mathrm{H}_2}\right]^{1 / 2}}{\left[\mathrm{H}^{+}\right]} \\ & =0-0.0591 \log \frac{1}{10^{-7}} \\ & =-0.0591 \log 10^7=-0.0591 \times 7 \log 10 \\ & =-0.0591 \times 7=-0.4137 \mathrm{~V} \end{aligned}$

Volume of $\mathrm{O}_2=5.6 \mathrm{~L}$

Number of moles of $\mathrm{O}_2=\frac{5.6}{22.4}=0.25 \mathrm{~mol}$

Equivalent of $\mathrm{O}_2=n$ factor $\times$ number of moles of $\mathrm{O}_2=4 \times 0.25=1$

Equivalent of $\mathrm{Cu}=$ equivalent of $\mathrm{O}_2=1$

$\frac{\text { Mass of } \mathrm{Cu}}{\text { Equivalent mass }}=1 \Rightarrow \frac{\text { Mass of } \mathrm{Cu}}{63.5 / 2}=1$

Mass of $\mathrm{Cu}=\frac{63.5}{2}=31.75 \mathrm{~g}$

$\begin{aligned} \mathrm{pH}_{\mathrm{l}} & =3 \\ \Rightarrow\left[\mathrm{H}^{+}\right]_1 & =10^{-3} \mathrm{M} \\ \mathrm{pH}_2 & =6 \\ \Rightarrow\left[\mathrm{H}^{+}\right]_2 & =10^{-6} \mathrm{M} \\ \text { Now, } E_{\text {cell }} & =E \Upsilon_{\text {cell }}+\frac{0.059}{1} \log \frac{10^{-3}}{10^{-6}}=0+0.059 \times 3 \\ E_{\text {cell }} & =0.177 \mathrm{~V}\end{aligned}$

Specific conductivity of $\mathrm{BaSO}_4$ solution

$\kappa_{\mathrm{BaSO}_4}=3.648 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$

Specific conductivity of $\mathrm{H}_2 \mathrm{O}, \mathrm{\kappa}_{\mathrm{H}_2 \mathrm{O}}=1.25 \times 10^{-6} \mathrm{~ohm}^{-1} \mathrm{~cm}^{-1}$

$\text { Now, } \begin{aligned} \kappa_{\text {BaSO }_4} & =\kappa_{\text {BaSO }_4} \text { (solution) }-\kappa_{\mathrm{H}_2 \mathrm{O}} \\ & =3.648 \times 10^{-6}-1.25 \times 10^{-6} \\ & =2.398 \times 10^{-6} \mathrm{Scm}^{-1} \end{aligned}$

Formula used, $\lambda_{\mathrm{m}}^{\circ}=\frac{1000 \times \kappa}{S}$ ..... (i)

Here, $S=$ solubility

$\begin{aligned} & \left.\kappa=\text { kappa (specific conductivity of } \mathrm{BaSO}_4\right) \\ & \lambda_{\mathrm{m}}\left(\mathrm{BaSO}_4\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Ba}^{2+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{SO}_4^{2-}\right) \\ & (110+136.6) \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}=246.6 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \end{aligned}$

Put values in Eq (i) we get,

$\begin{aligned} & 246.6 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \\ &=\frac{1000 \times 2.398 \times 10^{-6} \mathrm{~S} \mathrm{~cm}^{-1}}{S} \\ & S=9.72 \times 10^{-6} \times 233=2.266 \times 10^{-3} \mathrm{gL}^{-1} \end{aligned}$

Hence, the solubility of $\mathrm{BaSO}_4$ at $291 \mathrm{~K}$ will be $2.266 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$

Anode $\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 e^{-}$ ..... (i)

Cathode $\mathrm{Fe}^{2+} \longrightarrow 2 e^{-}+\mathrm{Fe}$ ..... (ii)

Multiply 2 in Eq. (i) and 3 in Eq. (ii) we get,

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

$=-0.42-(-0.72 \mathrm{~V})=-0.42+0.72 \mathrm{~V}$

$E_{\text {cell }}^{\circ}=0.30 \mathrm{~V}$

Using Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Fe}^{2+}\right]^3}$

Here, $n=6\left(\right.$ For $\left.6 e^{-}\right)$

$\begin{aligned} & E_{\text {cell }}=0.30-\frac{0.0591}{6} \log \frac{(0.1)^2}{(0.1)^3} \\ & E_{\text {cell }}=0.30-0 \Rightarrow E_{\text {cell }}=0.30 \mathrm{~V} \end{aligned}$

Hence, emf of the reaction cell is $0.30 \mathrm{~V}$.

$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$

Using Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Cr}^{3+}\right]^2\left[\mathrm{H}_2 \mathrm{O}\right]^7}{\left[\left(\mathrm{Cr}_2 \mathrm{O}_7\right)^{2-}\right]\left[\mathrm{H}^{+}\right]^{14}}$

$\begin{aligned} & 1.067=1.33-\frac{0.0591}{6} \log \frac{\left(1.5 \times 10^{-3}\right)^2(1)^7}{\left(4.5 \times 10^{-3}\right)\left[\mathrm{H}^{+}\right]^{14}} \\ & 0.26=\frac{0.0591}{6} \log \frac{\left[225 \times 10^{-6}\right]}{\left(4.5 \times 10^{-3}\right) \times\left[\mathrm{H}^{+}\right]^{14}} \\ & 26.7=\log \frac{50 \times 10^{-3}}{\left[\mathrm{H}^{+}\right]^{14}} \\ & \begin{array}{l}26.7=\log \left(50 \times 10^{-3}\right)-\log \left[\mathrm{H}^{+}\right]^{14} \\ \text { For } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ 26.7=-1.3010-14 \log \left[\mathrm{H}^{+}\right] \\ 14 \mathrm{pH}=28.041 \Rightarrow \mathrm{pH}=\frac{28.041}{14} \\ \mathrm{pH}=2\end{array}\end{aligned}$

Kolbe's electrolysis

$2 \mathrm{RCOOK}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow R-R+2 \mathrm{CO}_2+2 \mathrm{KOH}+\mathrm{H}_2$

Reaction at anode :

$\begin{aligned} 2 \mathrm{CH}_3 \mathrm{CO\dot O} & \longrightarrow 2 \dot{\mathrm{CH}}_3+2 \mathrm{CO}_2 \\ 2 \dot{\mathrm{H}}_3 & \longrightarrow \mathrm{C}_2 \mathrm{H}_6\end{aligned}$

As after completion of the reaction solution becomes alkaline, so $\mathrm{pH}$ of solution will increase. Sodium acetate on Kolbe's electrolysis gives ethane. It is formed at anode. $\dot{\mathrm{CH}}_3$ methyl radical is produced at anode only.

Hence, A is true but R is false.

Given current, $i=10 \mathrm{~A}$

Time, $t=1.608 \mathrm{~min}$ or $96.5 \mathrm{~s}$

Weight of aluminium $W_{\mathrm{Al}}=$ ?

Atomic weight of aluminium $=27 \mathrm{~g} / \mathrm{mol}(M)$

We know that, $\mathrm{AlCl}_3 \rightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-}, n=3$

At cathode, reduction $\mathrm{Al}^{3+}+3 e^{-} \rightarrow \mathrm{Al}$

Formula, $\quad W_{\mathrm{Al}}=\frac{\varepsilon \times i \times t}{96500}$, where $\varepsilon=\frac{M}{n}$

$\begin{aligned} & W_{\mathrm{Al}}=\frac{27}{3} \times \frac{10 \times 96.5}{96500} \\ & W_{\text {Al }}=0.09 \mathrm{~g} \end{aligned}$

Mass of $\mathrm{Al}$ deposited will be $0.09 \mathrm{~g}$

At infinite dilution, the molar conductivity of an electrolyte can be represented as the sum individual contribution of its cations and anions.

$\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_3\right) \longrightarrow \lambda_{\infty}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\infty}^{\circ}\left(\mathrm{NH}_2^{-}\right) \\ & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_3\right)=\lambda_{\infty}^{\circ}\left(\mathrm{NH}_2^{-}\right)+\lambda_{\infty}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda^{\circ} \\ & \mathrm{KBr}-\lambda^{\circ} \mathrm{KBr} \\ &=\lambda_{\infty}^{\circ}\left(\mathrm{KNH}_2\right)+\lambda_{\infty}^{\circ}(\mathrm{HBr})-\lambda_{\infty}^{\circ}(\mathrm{KBr}) \\ &=90.48+420.6-120.5 \\ &=390.5 \mathrm{~S} \mathrm{~cm} \mathrm{~mol}^{-1} . \end{aligned}$