Consider the following standard electrode potentials and calculate the eqillibrium constant at `25^(@)`C for the indicated disproportional reaction :
`3 Mn^(2+)(aq)toMn(s)+2Mn^(3+)(aq)`
`Mn^(3+)(aq)+e^(-)toMn^(2+)(aq), E^(@)=1.51 V`
`Mn^(2+)(aq)+2e^(-)toMn(s), E^(@)=-1.185 V``

Options

A. `1.2xx10^(-43)`
B. `2.4xx10^(-73)`
C. `6.3xx10^(-92)`
D. `1.5xx10^(-62)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions
The given disproportionation reaction is:
\[ 3 \text{Mn}^{2+}(aq) \rightarrow \text{Mn}(s) + 2 \text{Mn}^{3+}(aq) \]

We can break this down into two half-reactions:
1. Reduction:
\[ \text{Mn}^{3+}(aq) + e^- \rightarrow \text{Mn}^{2+}(aq) \quad (E^\circ = 1.51 \, \text{V}) \]
2. Oxidation:
\[ \text{Mn}^{2+}(aq) \rightarrow \text{Mn}(s) + 2e^- \quad (E^\circ = -1.185 \, \text{V}) \]

### Step 2: Adjust the half-reactions
To balance the electrons in the overall reaction, we multiply the reduction half-reaction by 2:
\[ 2 \text{Mn}^{3+}(aq) + 2e^- \rightarrow 2 \text{Mn}^{2+}(aq) \]

Now the half-reactions are:
1. Reduction:
\[ 2 \text{Mn}^{3+}(aq) + 2e^- \rightarrow 2 \text{Mn}^{2+}(aq) \quad (E^\circ = 1.51 \, \text{V}) \]
2. Oxidation:
\[ 3 \text{Mn}^{2+}(aq) \rightarrow \text{Mn}(s) + 2e^- \quad (E^\circ = -1.185 \, \text{V}) \]

### Step 3: Calculate the standard cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) is given by:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Here, the cathode is the reduction half-reaction and the anode is the oxidation half-reaction:
\[ E^\circ_{\text{cell}} = 1.51 \, \text{V} - (-1.185 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 1.51 + 1.185 = 2.695 \, \text{V} \]

### Step 4: Use the Nernst equation
The Nernst equation relates the standard cell potential to the equilibrium constant:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c \]

Where:
- \( n \) is the number of moles of electrons transferred. In this case, \( n = 2 \).

At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K_c \]
\[ \frac{0.0591}{2} \log K_c = 2.695 \]

### Step 5: Solve for \( K_c \)
Rearranging gives:
\[ \log K_c = \frac{2.695 \times 2}{0.0591} \]
\[ \log K_c = \frac{5.39}{0.0591} \]
\[ \log K_c \approx 91.201 \]

Now, converting from logarithmic form to exponential form:
\[ K_c = 10^{-91.201} \]
\[ K_c \approx 6.3 \times 10^{-92} \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at \( 25^\circ C \) is:
\[ K_c \approx 6.3 \times 10^{-92} \]

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