What is the standard cell Potential `(E_("cell")^@)` for following cell reaction ?
`2Fe(s)+O_2(g)+2H_2O(l)hArr2F^(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:

\[ 2Fe(s) + O_2(g) + 2H_2O(l) \rightleftharpoons 2Fe^{2+}(aq) + 4OH^-(aq) \]

we will follow these steps:

### Step 1: Identify the half-reactions
From the overall cell reaction, we can identify the oxidation and reduction half-reactions.

- **Oxidation half-reaction (Anode)**:
\[ 2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^- \]
Here, iron (Fe) is oxidized from 0 to +2 oxidation state.

- **Reduction half-reaction (Cathode)**:
\[ O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) \]
Here, oxygen (O) is reduced from 0 to -2 oxidation state.

### Step 2: Write the standard reduction potentials
We are given the following standard reduction potentials:
- For the oxidation of iron:
\[ E^\circ(Fe^{2+}/Fe) = -0.44 \, V \]
- For the reduction of oxygen:
\[ E^\circ(O_2/H_2O/OH^-) = 0.4 \, V \]

### Step 3: Determine the standard cell potential
The standard cell potential \( E_{\text{cell}}^\circ \) can be calculated using the formula:
\[ E_{\text{cell}}^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In our case:
- The cathode is the reduction half-reaction (Oxygen), so \( E^\circ_{\text{cathode}} = 0.4 \, V \).
- The anode is the oxidation half-reaction (Iron), so \( E^\circ_{\text{anode}} = -0.44 \, V \).

Substituting the values:
\[ E_{\text{cell}}^\circ = 0.4 \, V - (-0.44 \, V) = 0.4 \, V + 0.44 \, V = 0.84 \, V \]

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