Given the listed standard electrode potentials, what is `E^(@)` for the cell:
`4BiO^+(aq)+3N_2H_5^+(aq)to4Bi(s)3N_2(g)+4H_2O)l)+7H^+(aq)`
`N_2(g)+5H^+(aq)+4e^(-)toN_2H_5^+(aq),E^(@)=+0.23V`
`BiO^+(aq)+2H^+(aq)+3e^(-)toBi(s)+H_2O(l),E^(@)=+0.32V`

Options

A. 0.55
B. 0.34
C. 1.88
D. 0.09

Answer

To find the standard electrode potential \( E^\circ \) for the given electrochemical cell reaction:

### Step 1: Identify the half-reactions and their standard electrode potentials
We have two half-reactions provided:

1. **Reduction of bismuth oxide**:
\[ \text{BiO}^+(aq) + 2H^+(aq) + 3e^- \rightarrow \text{Bi}(s) + H_2O(l), \quad E^\circ = +0.32 \, \text{V} \]

2. **Reduction of nitrogen gas**:
\[ N_2(g) + 5H^+(aq) + 4e^- \rightarrow N_2H_5^+(aq), \quad E^\circ = +0.23 \, \text{V} \]

### Step 2: Determine the oxidation and reduction processes
- The bismuth oxide is being reduced to bismuth, so this half-reaction will act as the cathode.
- The nitrogen species \( N_2H_5^+ \) is being oxidized to nitrogen gas, which means we need to reverse the half-reaction for nitrogen.

### Step 3: Reverse the nitrogen half-reaction for oxidation
Reversing the nitrogen half-reaction gives:
\[ N_2H_5^+(aq) \rightarrow N_2(g) + 5H^+(aq) + 4e^-, \quad E^\circ = -0.23 \, \text{V} \]

### Step 4: Calculate the standard cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{BiO}^+/\text{Bi}} - E^\circ_{N_2H_5^+/N_2} \]
\[ E^\circ_{\text{cell}} = 0.32 \, \text{V} - (-0.23 \, \text{V}) = 0.32 \, \text{V} + 0.23 \, \text{V} = 0.55 \, \text{V} \]

### Final Answer
Thus, the standard electrode potential \( E^\circ \) for the cell is:
\[ \boxed{0.55 \, \text{V}} \]