Use the following standard electrode potentials, calculate `DeltaG^(@)` in kJ`//`mol for the indicated reaction :
`5Ce^(4+)(aq)+Mn^(2+)(aq)+4H_(2)O(l)to5Ce^(3+)(aq)+MnO_(4)^(-)(aq)+8H^(+)(aq)`
`MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51 V`
`Ce^(4+)(aq)+e^(=)toCe^(3+)(aq)" "E^(@)=+1.61 V`

Options

A. `-9.65`
B. `-24.3`
C. `-48.25`
D. `-35.2`

Answer

To calculate the standard Gibbs free energy change (ΔG°) for the given reaction, we will follow these steps:

### Step 1: Write the overall reaction
The reaction given is:
\[ 5 \text{Ce}^{4+} (aq) + \text{Mn}^{2+} (aq) + 4 \text{H}_2\text{O} (l) \rightarrow 5 \text{Ce}^{3+} (aq) + \text{MnO}_4^{-} (aq) + 8 \text{H}^{+} (aq) \]

### Step 2: Identify the half-reactions
From the problem, we have the following half-reactions:
1. **Reduction half-reaction for Cerium:**
\[ \text{Ce}^{4+} + e^{-} \rightarrow \text{Ce}^{3+} \quad E° = +1.61 \, \text{V} \]

2. **Reduction half-reaction for Manganese:**
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \quad E° = +1.51 \, \text{V} \]

### Step 3: Determine the oxidation and reduction processes
- **Cerium** is being reduced from +4 to +3, which involves gaining 5 electrons.
- **Manganese** is being oxidized from +2 to +7, which means we need to reverse the reduction half-reaction for manganese:
\[ \text{Mn}^{2+} \rightarrow \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 e^{-} \quad E° = -1.51 \, \text{V} \]

### Step 4: Calculate the cell potential (E°cell)
The overall cell potential (E°cell) is given by:
\[ E°_{cell} = E°_{reduction} + E°_{oxidation} \]
Substituting the values:
\[ E°_{cell} = 1.61 \, \text{V} + (-1.51 \, \text{V}) \]
\[ E°_{cell} = 0.10 \, \text{V} \]

### Step 5: Calculate ΔG°
Using the formula:
\[ ΔG° = -nFE°_{cell} \]
Where:
- \( n = 5 \) (number of electrons transferred)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( E°_{cell} = 0.10 \, \text{V} \)

Substituting these values:
\[ ΔG° = -5 \times 96500 \times 0.10 \]
\[ ΔG° = -48250 \, \text{J/mol} \]

### Step 6: Convert to kJ/mol
To convert Joules to kilojoules:
\[ ΔG° = -48.25 \, \text{kJ/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change for the reaction is:
\[ ΔG° = -48.25 \, \text{kJ/mol} \]

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