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
- $E^\circ_{\text{cell}} = -0.48$ V
- $E^\circ_{\text{cell}} = -0.04$ V
- $E^\circ_{\text{cell}} = +0.84$ V
- $E^\circ_{\text{cell}} = +1.28$ V
The Correct Option is C
Solution and Explanation
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