The standard oxidation potentials for `Mn^(3+)` ion acid solution are `Mn^(2+)overset(-1.5V)toMn^(3+)overset(-1.0V)toMnO_(2)`. Is the reaction `2Mn^(3+)+2H_(2)OtoMn^(2+)+MnO_(2)+4H^(+)` spontaneous under conditions of unit activity? What is the change in free energy?

Answer

To determine if the reaction \( 2Mn^{3+} + 2H_2O \rightarrow Mn^{2+} + MnO_2 + 4H^+ \) is spontaneous under conditions of unit activity and to calculate the change in free energy (\( \Delta G \)), we can follow these steps:

### Step 1: Identify the half-reactions and their standard potentials
The half-reactions and their standard potentials provided are:
1. \( Mn^{2+} + 2e^- \rightarrow Mn^{3+} \) with \( E^\circ = -1.5 \, V \)
2. \( Mn^{3+} + 2e^- \rightarrow MnO_2 + 4H^+ \) with \( E^\circ = -1.0 \, V \)

### Step 2: Write the overall reaction
The overall reaction can be derived from the half-reactions:
- The reverse of the first half-reaction (oxidation) gives:
\[ Mn^{3+} \rightarrow Mn^{2+} + 2e^- \]
- The second half-reaction (reduction) is:
\[ Mn^{3+} + 2e^- \rightarrow MnO_2 + 4H^+ \]

### Step 3: Combine the half-reactions
To combine the half-reactions, we need to ensure that the electrons cancel out. The overall reaction becomes:
\[ 2Mn^{3+} + 2H_2O \rightarrow Mn^{2+} + MnO_2 + 4H^+ \]

### Step 4: Calculate the standard cell potential \( E^\circ \)
The standard cell potential \( E^\circ \) for the overall reaction can be calculated as follows:
\[ E^\circ = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ = (-1.0 \, V) - (-1.5 \, V) = -1.0 + 1.5 = 0.5 \, V \]

### Step 5: Calculate the change in free energy \( \Delta G \)
The change in free energy can be calculated using the formula:
\[ \Delta G = -nFE^\circ \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 for this reaction)
- \( F \) = Faraday's constant \( \approx 96500 \, C/mol \)
- \( E^\circ \) = standard cell potential calculated in the previous step

Substituting the values:
\[ \Delta G = -2 \times 96500 \, C/mol \times 0.5 \, V = -48250 \, J/mol \]

### Conclusion
The reaction is spontaneous because \( \Delta G < 0 \). The change in free energy is \( -48250 \, J/mol \).

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