If the half cel reactions are given as
(i) `Fe^(2+)(Aq)+2e^(-)rarr Fe(s),E^(@)=-0.44 V`
(ii) `2H^(+)(sq)+1/2 O_(2)(g)+2e^(-)rarrH_(2)O(l) E^(@)=+1.23 V`
The `E^(@)` for the reaction
`Fe(s)+2H^(+)+1/2O_(2)(g)rarrFe^(2+)(aq)+H_(2)O(l)` will be

Options

A. `-0.79`
B. `-1.67 V`
C. `1.67 V`
D. `0.79 V`

Answer

To find the standard cell potential \( E^\circ \) for the reaction

\[ \text{Fe(s) + 2H}^+ + \frac{1}{2}\text{O}_2(g) \rightarrow \text{Fe}^{2+}(aq) + \text{H}_2\text{O}(l) \]

we will use the provided half-cell reactions and their standard potentials.

### Step 1: Identify the half-reactions and their standard potentials

We have the following half-cell reactions:

1. \( \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \) with \( E^\circ = -0.44 \, \text{V} \)
2. \( 2\text{H}^+(aq) + \frac{1}{2}\text{O}_2(g) + 2e^- \rightarrow \text{H}_2\text{O}(l) \) with \( E^\circ = +1.23 \, \text{V} \)

### Step 2: Determine which half-reaction is the anode and which is the cathode

In the overall reaction, iron is being oxidized (losing electrons) and hydrogen ions are being reduced (gaining electrons). Therefore:

- The oxidation half-reaction (anode) is:
\[ \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^- \]
This is the reverse of the first half-reaction, so we will take the negative of its standard potential:
\[ E^\circ_{\text{anode}} = +0.44 \, \text{V} \]

- The reduction half-reaction (cathode) is:
\[ 2\text{H}^+(aq) + \frac{1}{2}\text{O}_2(g) + 2e^- \rightarrow \text{H}_2\text{O}(l) \]
This retains its standard potential:
\[ E^\circ_{\text{cathode}} = +1.23 \, \text{V} \]

### Step 3: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)

We use the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 1.23 \, \text{V} - 0.44 \, \text{V} = 0.79 \, \text{V} \]

### Step 4: Conclusion

The standard cell potential \( E^\circ \) for the reaction is:
\[ E^\circ = 0.79 \, \text{V} \]