Question

Given below are half-cell reactions:
\(MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O\),
\(E°_{(Mn^{+2}/MnO_4^-)} = –1.510 V\)
\((\frac{1}{2})O_2 + 2H^+ + 2e^- → H_2O,\)
\(E°(O_2/H_2O) = +1.223 V\)
Will the permanganate ion, \(MnO_4^-\) liberate \(O_2\) from water in the presence of an acid?

• A. Yes, because E°cell = +0.287V
• B. No, because E°cell = -0.287V
• C. Yes, because E°cell = +2.733 V
• D. No, because E°cell = -2.733 V

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a reaction to be spontaneous and liberate a gas, the standard cell potential ($E^\circ_{cell}$) must be positive.
The liberation of $O_2$ from water implies that water is being oxidized to oxygen gas.
This must be coupled with the reduction of $MnO_4^-$ to $Mn^{2+}$.
Key Formula or Approach:
The standard cell potential is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Alternatively, \[ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \]
Step 2: Detailed Explanation:
1. Identify the Half-Reactions:
Reduction (Cathode): $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$; $E^\circ_{red} = +1.510$ V.
Oxidation (Anode): $H_2O \rightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$;
The standard reduction potential for oxygen is given as $E^\circ_{O_2/H_2O} = +1.223$ V.
The oxidation potential for water is the negative of this: $E^\circ_{ox} = -1.223$ V.

2. Calculate $E^\circ_{cell$:}
\[ E^\circ_{cell} = E^\circ_{MnO_4^-/Mn^{2+}} - E^\circ_{O_2/H_2O} \]
\[ E^\circ_{cell} = 1.510 \text{ V} - 1.223 \text{ V} \]
\[ E^\circ_{cell} = +0.287 \text{ V} \]

3. Feasibility:
Since $E^\circ_{cell}>0$, the reaction is thermodynamically spontaneous.
Therefore, $MnO_4^-$ will oxidize water to liberate $O_2$ in an acidic medium.
Step 3: Final Answer:
The reaction is feasible with $E^\circ_{cell} = +0.287$ V.