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. `-65.62 J` mole
B. `-196.86 kJ` mole
C. `- 98.43 kJ` mole
D. `-96.86 J `mole

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the reaction \(2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)\), we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials
The given standard electrode potentials are:
- For the reduction of chromium:
\[ Cr^{3+} + 3e^- \rightarrow Cr \quad (E^\circ = -0.74 \, V) \]
- For the reduction of cadmium:
\[ Cd^{2+} + 2e^- \rightarrow Cd \quad (E^\circ = -0.40 \, V) \]

### Step 2: Write the oxidation and reduction half-reactions
1. **Oxidation half-reaction** (for chromium):
\[ 2Cr \rightarrow 2Cr^{3+} + 6e^- \]
(This is the reverse of the reduction reaction, hence the sign of \(E^\circ\) changes.)

2. **Reduction half-reaction** (for cadmium):
\[ 3Cd^{2+} + 6e^- \rightarrow 3Cd \]

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Here, the cathode is the reduction of cadmium and the anode is the oxidation of chromium:
\[ E^\circ_{cell} = (-0.40 \, V) - (-0.74 \, V) = -0.40 + 0.74 = 0.34 \, V \]

### Step 4: Calculate the number of electrons transferred (n)
From the balanced equation, we see that 6 electrons are transferred in the overall reaction.

### Step 5: Use the Gibbs free energy equation
The relationship between Gibbs free energy change and cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \(n = 6\) (number of moles of electrons)
- \(F = 96500 \, C/mol\) (Faraday's constant)
- \(E^\circ_{cell} = 0.34 \, V\)

Substituting the values:
\[ \Delta G^\circ = -6 \times 96500 \, C/mol \times 0.34 \, V \]
\[ \Delta G^\circ = -6 \times 96500 \times 0.34 = -196.86 \, kJ/mol \]

### Step 6: Final answer
Thus, the standard Gibbs free energy change for the reaction is:
\[ \Delta G^\circ = -196.86 \, kJ/mol \]