Question:
What is the cell potential (standard emf, $E^\circ$) for the reaction below?
\[
E^\circ(\text{Fe}^{2+}/\text{Fe}) = 0.44\text{ V}, \quad E^\circ(\text{O}_2/\text{H}_2\text{O}/\text{OH}^-) = 0.40\text{ V}
\]
\[
2\text{Fe}(s) + \text{O}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{Fe}^{2+}(aq) + 4\text{OH}^-(aq)
\]
What is the cell potential (standard emf, $E^\circ$) for the reaction below?
\[
E^\circ(\text{Fe}^{2+}/\text{Fe}) = 0.44\text{ V}, \quad E^\circ(\text{O}_2/\text{H}_2\text{O}/\text{OH}^-) = 0.40\text{ V}
\]
\[
2\text{Fe}(s) + \text{O}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{Fe}^{2+}(aq) + 4\text{OH}^-(aq)
\]
Show Hint
Use reduction potentials; reverse sign for oxidation.
Updated On: Apr 23, 2026
- $E^\circ_{\text{cell}} = -0.48$ V
- $E^\circ_{\text{cell}} = -0.04$ V
- $E^\circ_{\text{cell}} = +0.84$ V
- $E^\circ_{\text{cell}} = +1.28$ V
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\]
Step 1: Fe $\rightarrow$ Fe$^{2+}$ = oxidation (anode).
Step 2: O$_2$ $\rightarrow$ OH$^-$ = reduction (cathode).
Step 3: Substitute values: \[ E^\circ_{\text{cell}} = \textcolor{red}{0.40 - (-0.44)} \]
Step 4: \[ E^\circ_{\text{cell}} = 0.40 + 0.44 = \textcolor{red}{0.84\text{ V}} \] Conclusion:
Cell potential = +0.84 V
Step 1: Fe $\rightarrow$ Fe$^{2+}$ = oxidation (anode).
Step 2: O$_2$ $\rightarrow$ OH$^-$ = reduction (cathode).
Step 3: Substitute values: \[ E^\circ_{\text{cell}} = \textcolor{red}{0.40 - (-0.44)} \]
Step 4: \[ E^\circ_{\text{cell}} = 0.40 + 0.44 = \textcolor{red}{0.84\text{ V}} \] Conclusion:
Cell potential = +0.84 V
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