The standard rectuion potentials for two half-cell reactions are given below
`Cd^(2+)(aq)+2e^(-)rarrCd(s),E^(@)=-0.40V`
`Ag^(+)(aq)+e^(-)rarrAg(s),E^(@)=0.80V`
The standard free energy change for the reaction
`2Ag^(+)(aq)+Cd(s)rarr2Ag(s)+Cd^(2+)(aq)` is given by

Options

A. `115.8kJ`
B. `-115.8kJ`
C. `-231.6kJ`
D. `231.6kJ`

Explanation

To find the standard free energy change (ΔG°) for the reaction:

\[ 2Ag^+(aq) + Cd(s) \rightarrow 2Ag(s) + Cd^{2+}(aq) \]

we will follow these steps:

### Step 1: Identify the Half-Reactions
The given half-reactions are:
1. Reduction of silver ion:
\[ Ag^+(aq) + e^- \rightarrow Ag(s), \quad E^\circ = 0.80 \, V \]
2. Reduction of cadmium ion:
\[ Cd^{2+}(aq) + 2e^- \rightarrow Cd(s), \quad E^\circ = -0.40 \, V \]

### Step 2: Write the Oxidation Half-Reaction
Since cadmium is being oxidized in the overall reaction, we need to reverse the reduction half-reaction for cadmium:
\[ Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-, \quad E^\circ = +0.40 \, V \]

### Step 3: Combine Half-Reactions
Now we can combine the oxidation and reduction half-reactions:
- Oxidation (anode):
\[ Cd(s) \rightarrow Cd^{2+}(aq) + 2e^- \]
- Reduction (cathode):
\[ 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \]

Adding these gives the overall cell reaction:
\[ 2Ag^+(aq) + Cd(s) \rightarrow 2Ag(s) + Cd^{2+}(aq) \]

### Step 4: Calculate the Standard Cell Potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0.80 \, V - (-0.40 \, V) = 0.80 \, V + 0.40 \, V = 1.20 \, V \]

### Step 5: Calculate the Standard Free Energy Change (ΔG°)
The relationship between the standard free energy change and the standard cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred (2 for this reaction)
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E^\circ_{cell} = 1.20 \, V \)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 1.20 \, V \]
\[ \Delta G^\circ = -231600 \, J/mol \]
Converting to kilojoules:
\[ \Delta G^\circ = -231.6 \, kJ/mol \]

### Final Answer
The standard free energy change for the reaction is:
\[ \Delta G^\circ = -231.6 \, kJ/mol \]

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