Under standered condition `Delta G^(@)` for the reaction `2Cr(s)= 3Cd^(2+)(aq) rarr 2Cr_((a a))^(3+) +3Cd(s)` is
`(E_(Cr^(3+)//Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.4 V)`

Options

A. `-196.86` kJ/mol
B. `-98.43` kJ/mol
C. `-65.62` kJ/mol
D. `-164.05` kJ/mol

Answer

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the given reaction using the standard electrode potentials provided. Here’s a step-by-step solution:

### Step 1: Identify the Reaction
The reaction given is:
\[ 2 \text{Cr(s)} + 3 \text{Cd}^{2+}(aq) \rightarrow 2 \text{Cr}^{3+}(aq) + 3 \text{Cd(s)} \]

### Step 2: Determine Oxidation and Reduction
In this reaction:
- Chromium (Cr) is oxidized from Cr(s) to Cr³⁺(aq).
- Cadmium (Cd²⁺) is reduced from Cd²⁺(aq) to Cd(s).

### Step 3: Write Half-Reactions
The half-reactions can be written as:
- Oxidation (at anode):
\[ \text{Cr(s)} \rightarrow \text{Cr}^{3+}(aq) + 3e^- \]
- Reduction (at cathode):
\[ \text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd(s)} \]

### Step 4: Calculate the Number of Electrons Transferred (n)
From the balanced half-reactions, we can see that:
- 2 moles of Cr produce 6 electrons (since each Cr gives 3 electrons).
- 3 moles of Cd²⁺ consume 6 electrons (since each Cd²⁺ requires 2 electrons).

Thus, \( n = 6 \).

### Step 5: Find Standard Electrode Potentials
Given:
- \( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V} \)
- \( E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, \text{V} \)

### Step 6: Calculate the Standard Cell Potential (E°cell)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, Cd is reduced (cathode) and Cr is oxidized (anode):
\[ E^\circ_{\text{cell}} = (-0.40 \, \text{V}) - (-0.74 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.40 + 0.74 = 0.34 \, \text{V} \]

### Step 7: Calculate ΔG°
Using the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( n = 6 \)
- \( E^\circ_{\text{cell}} = 0.34 \, \text{V} \)

Substituting the values:
\[ \Delta G^\circ = -6 \times 96500 \, \text{C/mol} \times 0.34 \, \text{V} \]
\[ \Delta G^\circ = -6 \times 96500 \times 0.34 \]
\[ \Delta G^\circ = -196860 \, \text{J/mol} \]

### Step 8: Convert to kJ/mol
\[ \Delta G^\circ = -196.86 \, \text{kJ/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change (ΔG°) for the reaction is:
\[ \Delta G^\circ = -196.86 \, \text{kJ/mol} \]