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

Options

A. spontaneous ,`-48250J`
B. nonspontaneous, `+48250J`
C. no change in free energy
D. spontaneous , `-96500J`

Explanation

To determine if the reaction \(2Mn^{3+} + 2H_2O \rightarrow Mn^{2+} + MnO_2 + 4H^+\) is spontaneous under conditions of unit activity, we will calculate the change in free energy (\(\Delta G\)) using the standard reduction potentials provided.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials**:
- The standard reduction potential for \(Mn^{3+} + e^- \rightarrow Mn^{2+}\) is \(E^\circ = -1.5 \, V\).
- The standard reduction potential for \(MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{3+} + 2H_2O\) can be derived from the given potentials. Since \(Mn^{3+}\) is being reduced, we will consider the reverse reaction for \(Mn^{3+}\) to \(Mn^{2+}\), which gives us \(E^\circ = +1.5 \, V\).

2. **Calculate the Standard Cell Potential**:
- The overall cell reaction involves the reduction of \(Mn^{3+}\) and the oxidation of \(Mn^{2+}\) to \(MnO_2\).
- The standard cell potential (\(E^\circ_{cell}\)) can be calculated as:
\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \]
- For our reaction:
- Reduction: \(Mn^{3+} + e^- \rightarrow Mn^{2+}\) (which we take as \(+1.5 \, V\))
- Oxidation: \(MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{3+} + 2H_2O\) (which we take as \(-1.0 \, V\))
- Therefore:
\[ E^\circ_{cell} = 1.5 - (-1.0) = 1.5 + 1.0 = 2.5 \, V \]

3. **Calculate the Change in Free Energy**:
- The change in free energy (\(\Delta G^\circ\)) can be calculated using the equation:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
- Where:
- \(n\) = number of moles of electrons transferred (from the balanced equation, \(n = 2\))
- \(F\) = Faraday's constant (\(96500 \, C/mol\))
- \(E^\circ_{cell} = 2.5 \, V\)
- Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \times 2.5 \]
\[ \Delta G^\circ = -482500 \, J \quad \text{(or -482.5 kJ)} \]

4. **Determine Spontaneity**:
- A negative \(\Delta G^\circ\) indicates that the reaction is spontaneous under standard conditions.

### Final Answer:
- The reaction \(2Mn^{3+} + 2H_2O \rightarrow Mn^{2+} + MnO_2 + 4H^+\) is spontaneous.
- The change in free energy is \(\Delta G^\circ = -482500 \, J\) or \(-482.5 \, kJ\).