Q.Which of the following statement is correct?

The key idea is that a galvanic cell converts chemical energy into electrical work. The standard Gibbs free energy change $\Delta_r G^\Theta$ is directly related to the maximum electrical work the cell can do, which is given by the product of the charge transferred and the cell potential.

We are given the overall reaction:

$2\text{Cr}(s) + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd}(s)$

To find $\Delta_r G^\Theta$, we need the standard cell potential $E^\Theta_\text{cell}$ and the number of electrons transferred $n$.

1. Identify the half-reactions and their standard potentials

   • Reduction of cadmium: $\text{Cd}^{2+} + 2e^- \rightarrow \text{Cd}(s)$, $E^\circ = -0.40\ \text{V}$
   • Reduction of chromium: $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}(s)$, $E^\circ = -0.74\ \text{V}$

   In the overall reaction, $\text{Cr}$ is oxidized (it loses electrons) and $\text{Cd}^{2+}$ is reduced. So the cell potential is:

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

   The cathode is where reduction occurs: $\text{Cd}^{2+}|\text{Cd}$ with $E^\circ = -0.40\ \text{V}$

   The anode is where oxidation occurs: $\text{Cr}|\text{Cr}^{3+}$ with $E^\circ = -0.74\ \text{V}$ (but we use the reduction potential as given, then subtract).

   Thus:

   $E^\Theta_\text{cell} = (-0.40) - (-0.74) = +0.34\ \text{V}$

2. Determine the number of electrons transferred ($n$)

   From the half-reactions:

   • $\text{Cd}^{2+} + 2e^- \rightarrow \text{Cd}$ (each Cd²⁺ takes 2 electrons)
   • $\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-$ (each Cr loses 3 electrons)

   To balance the overall reaction, we need the least common multiple of 2 and 3, which is 6.

   Multiply the cadmium half-reaction by 3: $3\text{Cd}^{2+} + 6e^- \rightarrow 3\text{Cd}$

   Multiply the chromium half-reaction by 2: $2\text{Cr} \rightarrow 2\text{Cr}^{3+} + 6e^-$

   So $n = 6$ moles of electrons transferred per mole of reaction as written.

3. Apply the relationship between $\Delta_r G^\Theta$ and $E^\Theta_\text{cell}$

   The fundamental equation is:

   $\Delta_r G^\Theta = -n F E^\Theta_\text{cell}$

   where $F = 96500\ \text{C mol}^{-1}$ (Faraday constant).

   Substitute:

   $\Delta_r G^\Theta = -6 \times 96500 \times 0.34$

   First compute $6 \times 96500 = 579000$

   Then $579000 \times 0.34 = 196860$

   So $\Delta_r G^\Theta = -196860\ \text{J mol}^{-1}$

   Convert to kJ mol⁻¹:

   $\Delta_r G^\Theta = -196.86\ \text{kJ mol}^{-1}$

4. Match with the options

   The value $-196.86$ corresponds to option (A).

The key concept here is the Nernst equation, which tells us how the cell potential changes when concentrations deviate from standard conditions (1 M for solutes, 1 atm for gases, pure solids ignored). For the reaction

the standard potential is $E^\circ = 1.05\ \text{V}$. The student wants $E > E^\circ$. Since $E^\circ$ is fixed, we must adjust the reaction quotient $Q$ to make $E$ larger.

1. Write the Nernst equation for this cell At 25°C, the Nernst equation is

where $n$ is the number of electrons transferred. Here, Ni loses 2 electrons and each Ag⁺ gains 1, so $n = 2$.

The reaction quotient is

(Solids like Ni and Ag do not appear in $Q$.)

2. How to make $E > E^\circ$? Since $E = E^\circ - \frac{0.0592}{2} \log Q$, we need $E > E^\circ$, which means

So the cell potential exceeds the standard value when the reaction quotient is less than 1.

3. What does $Q < 1$ mean in terms of concentrations?

Under standard conditions, both concentrations are 1 M, so $Q = 1$ and $E = E^\circ$. To make $Q < 1$, we can either decrease $[\text{Ni}^{2+}]$ or increase $[\text{Ag}^+]$.

4. Evaluate the options
   • (A) Increase the concentration of Ag⁺(aq) → This makes the denominator larger, so $Q$ becomes smaller → $E$ increases. ✓
   • (B) Increase the size of the Ni(s) electrode → Solids do not appear in $Q$; size has no effect on potential. ✗
   • (C) Decrease the size of the Ag(s) electrode → Same reasoning: solids are irrelevant to $Q$. ✗
   • (D) Increase the pressure → No gases are involved; pressure changes do not affect concentrations of dissolved ions significantly. ✗

Concept and intuition:

In a galvanic cell, the spontaneous reaction drives electrons from the anode (oxidation) to the cathode (reduction). The standard cell potential $E^\Theta_{\text{cell}}$ is the difference between the reduction potentials of the two half-cells. From $E^\Theta_{\text{cell}}$ we directly get the Gibbs free energy change via $\Delta_r G^\Theta = -nFE^\Theta_{\text{cell}}$, and the equilibrium constant via the Nernst equation at standard conditions: $\log K_c = \frac{nE^\Theta_{\text{cell}}}{0.059}$ at 298 K. The sign of $\Delta_r G^\Theta$ must be negative for a spontaneous reaction, which immediately eliminates options with positive $\Delta_r G^\Theta$.

Step-by-step reasoning:

1. Identify the half-reactions and their standard potentials.

   • Reduction: $\ce{Ag^+ + e^- -> Ag(s)}$, $E^\Theta = +0.80\ \text{V}$ (cathode).
   • Oxidation: $\ce{Fe^{2+} -> Fe^{3+} + e^-}$, but we have the reduction potential for $\ce{Fe^{3+}/Fe^{2+}} = +0.77\ \text{V}$. For oxidation, we reverse the sign: $E^\Theta_{\text{ox}} = -0.77\ \text{V}$. The cell potential is $E^\Theta_{\text{cell}} = E^\Theta_{\text{cathode}} - E^\Theta_{\text{anode}} = 0.80 - 0.77 = 0.03\ \text{V}$.

2. Determine the number of electrons transferred ($n$).

   The balanced reaction $\ce{Fe^{2+} + Ag^+ -> Fe^{3+} + Ag}$ involves one electron transfer: $\ce{Fe^{2+} -> Fe^{3+} + e^-}$ and $\ce{Ag^+ + e^- -> Ag}$. So $n = 1$.

3. Calculate $\Delta_r G^\Theta$.

   $\Delta_r G^\Theta = -nFE^\Theta_{\text{cell}} = -1 \times 96500\ \text{C mol}^{-1} \times 0.03\ \text{V}$.

   Since $1\ \text{C}\cdot\text{V} = 1\ \text{J}$, we get $\Delta_r G^\Theta = -2895\ \text{J mol}^{-1} = -2.895\ \text{kJ mol}^{-1}$.

4. Calculate $\log K_c$.

   At 298 K, the Nernst equation simplifies to $E^\Theta_{\text{cell}} = \frac{0.059}{n} \log K_c$ (using $R = 8.3$, $F = 96500$, and $\ln \to \log_{10}$ factor 2.303).

   Rearranging: $\log K_c = \frac{nE^\Theta_{\text{cell}}}{0.059} = \frac{1 \times 0.03}{0.059} \approx 0.5085 \approx 0.508$.

5. Match with the options.

   The pair $(-2.895,\ 0.508)$ corresponds to option (C).

The spontaneous reaction

has $\text{Zn}^{2+}$ reduced (cathode) and M oxidised (anode). Therefore

So the standard reduction potential of $\text{M}^{3+}+3e^- \rightarrow \text{M}$ is $-1.66$ V.

The trap here is tempting: just add $0.153\ \text{V}$ and $0.520\ \text{V}$ to get $0.673\ \text{V}$. That would be the answer if potentials were additive — but they are not. Electrode potentials are intensive properties; they depend on the number of electrons transferred. What is additive is the Gibbs free energy change, $\Delta G^\circ = -nFE^\circ$.

So to find $E^\circ$ for $\ce{Cu^{2+} + 2e^- -> Cu}$, we must work through the free energies.

1. Write the half-reactions with their $n$ and $\Delta G^\circ$.

   For $\ce{Cu^{2+} + e^- -> Cu^+}$:

   $n_1 = 1$, $E_1^\circ = +0.153\ \text{V}$

   $\Delta G_1^\circ = -n_1 F E_1^\circ = -F(0.153)$

   For $\ce{Cu^+ + e^- -> Cu}$:

   $n_2 = 1$, $E_2^\circ = +0.520\ \text{V}$

   $\Delta G_2^\circ = -F(0.520)$

2. Add the two steps to get the overall reaction.

   $\ce{Cu^{2+} + e^- -> Cu^+}$

   $\ce{Cu^+ + e^- -> Cu}$

   Sum: $\ce{Cu^{2+} + 2e^- -> Cu}$

   The overall $\Delta G^\circ$ is the sum:

   $\Delta G^\circ_{\text{total}} = \Delta G_1^\circ + \Delta G_2^\circ = -F(0.153 + 0.520) = -F(0.673)$

3. Relate total $\Delta G^\circ$ to the overall $E^\circ$.

   For the overall reaction, $n = 2$. So:

   $\Delta G^\circ_{\text{total}} = -n F E^\circ_{\ce{Cu^{2+}/Cu}} = -2F E^\circ_{\ce{Cu^{2+}/Cu}}$

   Equate:

   $-2F E^\circ_{\ce{Cu^{2+}/Cu}} = -F(0.673)$

   Cancel $-F$ (non-zero):

   $2 E^\circ_{\ce{Cu^{2+}/Cu}} = 0.673$

   Hence:

   $E^\circ_{\ce{Cu^{2+}/Cu}} = \frac{0.673}{2} = 0.3365\ \text{V} \approx 0.336\ \text{V}$

The key idea is that the standard Gibbs free energy change ($\Delta G^\circ$) for a spontaneous electrochemical cell reaction is negative and proportional to the cell potential. The relationship is $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, where $n$ is the number of moles of electrons transferred in the balanced reaction, and $F$ is Faraday’s constant (charge per mole of electrons). This formula comes from the fact that electrical work done by the cell equals the charge transferred times the potential difference, and at constant temperature and pressure, that work equals the decrease in Gibbs free energy.

1. Determine the number of electrons transferred ($n$).

   The half-reactions are:

   • Oxidation: $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
   • Reduction: $\text{Ni}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Ni(s)}$ Each zinc atom loses two electrons, and each nickel ion gains two electrons. So $n = 2$.

2. Apply the formula $\Delta G^\circ = -nFE^\circ_{\text{cell}}$.

   Given:

   $E^\circ_{\text{cell}} = 0.51\ \text{V}$

   $F = 96500\ \text{C mol}^{-1}$

   Therefore:

3. Calculate step by step.

   First, $2 \times 96500 = 193000$.

   Then, $193000 \times 0.51 = 98430$.

   So $\Delta G^\circ = -98430\ \text{J mol}^{-1}$.

4. Convert to kJ mol$^{-1}$.

   Since $1\ \text{kJ} = 1000\ \text{J}$:

Concept. We're given $\Delta U$ and $\Delta S$ for a reaction where a liquid converts into gas. Since gas is produced, the reaction does $P\Delta V$ work on the surroundings, so $\Delta H \ne \Delta U$. We first find $\Delta H$, then apply $\Delta G = \Delta H - T\Delta S$.

1. Relation between $\Delta H$ and $\Delta U$. At constant pressure, $\Delta H = \Delta U + P\Delta V = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in moles of gas.

2. Find $\Delta n_g$. Reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$. The reactant is a liquid (0 mol gas); the product is 2 mol gas. So $\Delta n_g = 2$.

3. Compute $P\Delta V$. With $R = 2~\mathrm{Cal\,K^{-1}mol^{-1}}$ and $T = 25^\circ\mathrm{C} = 298~\mathrm{K}$:

4. Compute $\Delta H$.

5. Compute $T\Delta S$. Converting $\Delta S = 20~\mathrm{Cal/K}$ to k.Cal/K: $\Delta S = 0.020~\mathrm{k.Cal/K}$.

6. Compute $\Delta G$.