What is the standard cell Potential `(E_("cell")^@)` for following cell reaction ?
`2Fe(s)+O_2(g)+2H_2O(l)hArr2Fe^(2+)(aq)+4OH^(-)(aq)`
Given
`E_(Fe^(2+)(aq)|Fe=-0.44V)^@`
`E_(O_2(g)|H_2O|OH=0.4V)^@`

Options

A. `E_("cell")^@=0.48V`
B. `E_("cell")^@=0.04V`
C. `E_("cell")^@=+0.84V`
D. `E_("cell")^@=+1.28V`

Answer

To find the standard cell potential \( E_{\text{cell}}^\circ \) for the given cell reaction:

\[ 2 \text{Fe}(s) + \text{O}_2(g) + 2 \text{H}_2\text{O}(l) \rightleftharpoons 2 \text{Fe}^{2+}(aq) + 4 \text{OH}^-(aq) \]

we will follow these steps:

### Step 1: Identify the half-reactions

1. **Oxidation half-reaction** (at the anode):
\[ 2 \text{Fe}(s) \rightarrow 2 \text{Fe}^{2+}(aq) + 4 e^- \]
The standard reduction potential for this reaction is given as:
\[ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \]

2. **Reduction half-reaction** (at the cathode):
\[ \text{O}_2(g) + 2 \text{H}_2\text{O}(l) + 4 e^- \rightarrow 4 \text{OH}^-(aq) \]
The standard reduction potential for this reaction is given as:
\[ E^\circ_{\text{O}_2/\text{OH}^-} = 0.4 \, \text{V} \]

### Step 2: Calculate the standard cell potential

The standard cell potential \( E_{\text{cell}}^\circ \) can be calculated using the formula:
\[ E_{\text{cell}}^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E_{\text{cell}}^\circ = E^\circ_{\text{O}_2/\text{OH}^-} - E^\circ_{\text{Fe}^{2+}/\text{Fe}} \]
\[ E_{\text{cell}}^\circ = 0.4 \, \text{V} - (-0.44 \, \text{V}) \]
\[ E_{\text{cell}}^\circ = 0.4 \, \text{V} + 0.44 \, \text{V} \]
\[ E_{\text{cell}}^\circ = 0.84 \, \text{V} \]

### Conclusion

Thus, the standard cell potential \( E_{\text{cell}}^\circ \) for the given cell reaction is:
\[ \boxed{0.84 \, \text{V}} \]