Question:
If \(E^\circ_{MnO_4^-/MnO_2} = x \, V\) and \(E^\circ_{MnO_2/Mn^{2+}} = y \, V\), what is the value of \(E^\circ_{MnO_4^-/Mn^{2+}}\)?
If \(E^\circ_{MnO_4^-/MnO_2} = x \, V\) and \(E^\circ_{MnO_2/Mn^{2+}} = y \, V\), what is the value of \(E^\circ_{MnO_4^-/Mn^{2+}}\)?
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Combine electrode potentials using electron-weighted averaging, not direct addition.
Updated On: Jun 20, 2026
- \( \frac{3x + 2y}{3} \)
- \( \frac{3x + 2y}{5} \)
- \( \frac{5x + 3y}{2} \)
- \( \frac{3x + 5y}{3} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understand concept of standard electrode potential.
When multiple redox steps are involved, overall potential is not simple addition but weighted based on electron transfer numbers in each step.
Step 2: Break into half reactions.
MnO\(_4^-\) → MnO\(_2\) corresponds to 3 electron change. MnO\(_2\) → Mn\(^{2+}\) corresponds to 2 electron change.
Step 3: Use energy additivity principle.
\[ \Delta G^\circ = -nFE^\circ \] Total free energy change is additive, so: \[ n_{total}E_{total} = n_1E_1 + n_2E_2 \]
Step 4: Substitute values.
Total electrons = 5: \[ 5E = 3x + 2y \]
Step 5: Solve for E.
\[ E = \frac{3x + 2y}{5} \]
Step 6: Final conclusion.
Thus standard potential for overall reaction is weighted average based on electrons transferred.
Final Answer: \[ \boxed{\frac{3x + 2y}{5}} \]
When multiple redox steps are involved, overall potential is not simple addition but weighted based on electron transfer numbers in each step.
Step 2: Break into half reactions.
MnO\(_4^-\) → MnO\(_2\) corresponds to 3 electron change. MnO\(_2\) → Mn\(^{2+}\) corresponds to 2 electron change.
Step 3: Use energy additivity principle.
\[ \Delta G^\circ = -nFE^\circ \] Total free energy change is additive, so: \[ n_{total}E_{total} = n_1E_1 + n_2E_2 \]
Step 4: Substitute values.
Total electrons = 5: \[ 5E = 3x + 2y \]
Step 5: Solve for E.
\[ E = \frac{3x + 2y}{5} \]
Step 6: Final conclusion.
Thus standard potential for overall reaction is weighted average based on electrons transferred.
Final Answer: \[ \boxed{\frac{3x + 2y}{5}} \]
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