Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell,
`Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "`

Explanation

To calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode, we can use the Nernst equation and the information given in the problem.

### Step 1: Write down the cell reaction and the Nernst equation.
The cell potential (E_cell) can be expressed using the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
Where:
- \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin (assume 298 K if not specified)
- \( n \) is the number of moles of electrons transferred in the reaction
- \( F \) is Faraday's constant (96485 C/mol)
- \( Q \) is the reaction quotient

### Step 2: Identify the components of the cell.
In the cell `Ni//Ni^(2+)(0.01 M)//Cu`, nickel is the anode and copper is the cathode. The standard reduction potential for copper is given as:
\[ E^\circ_{Cu^{2+}/Cu} = +0.34 \, V \]

### Step 3: Calculate the standard cell potential.
The overall cell potential can be expressed as:
\[ E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Ni^{2+}/Ni} \]

### Step 4: Use the given cell potential to find \( E^\circ_{Ni^{2+}/Ni} \).
We know:
- The cell potential \( E_{cell} = 0.59 \, V \)
- The standard reduction potential for copper \( E^\circ_{Cu^{2+}/Cu} = +0.34 \, V \)

Substituting into the equation:
\[ 0.59 = 0.34 - E^\circ_{Ni^{2+}/Ni} \]

### Step 5: Rearranging the equation to solve for \( E^\circ_{Ni^{2+}/Ni} \).
Rearranging gives:
\[ E^\circ_{Ni^{2+}/Ni} = 0.34 - 0.59 \]
\[ E^\circ_{Ni^{2+}/Ni} = -0.25 \, V \]

### Step 6: Final answer.
Thus, the standard electrode potential of the `Ni^(2+)//Ni` electrode is:
\[ E^\circ_{Ni^{2+}/Ni} = -0.25 \, V \]

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