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 cell reaction, we need to analyze the half-reactions involved and apply the formula for the overall cell potential. Here are the steps to solve the problem:

### Step 1: Identify the half-reactions
The overall cell reaction is:
\[ 4 \text{BiO}^+(aq) + 3 \text{N}_2\text{H}_5^+(aq) \rightarrow 4 \text{Bi}(s) + 3 \text{N}_2(g) + 4 \text{H}_2O(l) + 7 \text{H}^+(aq) \]

From the provided standard electrode potentials:
1. For the reduction of bismuth:
\[ \text{BiO}^+(aq) + 2 \text{H}^+(aq) + 3 e^- \rightarrow \text{Bi}(s) + \text{H}_2O(l), \quad E^\circ = +0.32 \, \text{V} \]
2. For the reduction of nitrogen:
\[ \text{N}_2(g) + 5 \text{H}^+(aq) + 4 e^- \rightarrow \text{N}_2\text{H}_5^+(aq), \quad E^\circ = -0.23 \, \text{V} \]

### Step 2: Determine oxidation and reduction
- **Reduction**: The bismuth species is being reduced from \( \text{BiO}^+ \) to \( \text{Bi} \).
- **Oxidation**: The nitrogen species \( \text{N}_2\text{H}_5^+ \) is being oxidized to \( \text{N}_2 \).

### Step 3: Reverse the oxidation half-reaction
To find the oxidation potential, we need to reverse the nitrogen half-reaction:
\[ \text{N}_2\text{H}_5^+(aq) \rightarrow \text{N}_2(g) + 5 \text{H}^+(aq) + 4 e^-, \quad E^\circ = +0.23 \, \text{V} \]

### Step 4: Calculate the overall cell potential
Now, we can calculate the overall cell potential using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (+0.32 \, \text{V}) + (+0.23 \, \text{V}) = +0.55 \, \text{V} \]

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

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