The cell potential for the cell
`Ni|Ni^(2+)(1M)||Cu^(2+)(1M)|Cu` is 0.59 V and `E(Cu^(2+)|Cu) = +0.34 V`. What is the standard electrode potential of `Ni^(2+)|Ni` electrode?

Options

A. `-0.25 V`
B. `+0.25 V`
C. `+9.93 V`
D. `-0.93 V`

Explanation

To find the standard electrode potential of the Ni²⁺|Ni electrode, we can follow these steps:

### Step 1: Understand the Cell Notation
The cell notation given is:
\[ \text{Ni} | \text{Ni}^{2+} (1M) || \text{Cu}^{2+} (1M) | \text{Cu} \]
This indicates that nickel is the anode (oxidation occurs) and copper is the cathode (reduction occurs).

### Step 2: Write the Half-Reactions
- At the anode (oxidation):
\[ \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \]

- At the cathode (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 3: Overall Cell Reaction
Combining the half-reactions gives the overall cell reaction:
\[ \text{Ni} + \text{Cu}^{2+} \rightarrow \text{Ni}^{2+} + \text{Cu} \]

### Step 4: Use the Nernst Equation
The cell potential \( E_{\text{cell}} \) is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \( E_{\text{cell}} = 0.59 \, \text{V} \)
- \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \)
- \( n = 2 \) (number of electrons transferred)

### Step 5: Substitute Known Values
We know:
- \( E^{\circ}_{\text{cathode}} = E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V} \)
- Concentrations of both ions are 1M, hence the logarithmic term becomes \( \log(1) = 0 \).

Thus, the equation simplifies to:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} \]
So,
\[ 0.59 \, \text{V} = E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} - E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} \]

### Step 6: Rearranging to Find \( E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} \)
Substituting the known values:
\[ 0.59 \, \text{V} = 0.34 \, \text{V} - E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} \]

Rearranging gives:
\[ E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = 0.34 \, \text{V} - 0.59 \, \text{V} \]
\[ E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \]

### Final Answer
The standard electrode potential of the Ni²⁺|Ni electrode is:
\[ E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \]

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