The solution of nickel sulphate in which nickel rod is dipped is diluted 10 times. The halfcell potential of nickel: 

Options

A. Decreases by 60 mV
B. Increases by 30 m V
C. Decreases by 30 m V
D. Decreases by 60 V

Answer

To solve the problem of finding the half-cell potential of a nickel electrode when the nickel sulfate solution is diluted 10 times, we will use the Nernst equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Nernst Equation
The Nernst equation relates the cell potential (E) to the standard electrode potential (E°) and the concentrations of the reactants and products. The equation is given by:

\[ E = E° - \frac{0.059}{n} \log \frac{[\text{oxidized form}]}{[\text{reduced form}]} \]

For nickel, the half-reaction can be represented as:

\[ \text{Ni}^{2+} + 2e^- \leftrightarrow \text{Ni} \]

Here, \( n = 2 \) because 2 electrons are involved in the reduction of nickel ions.

### Step 2: Set Up the Initial Conditions
Let’s denote the initial concentration of nickel ions in the solution as \( [\text{Ni}^{2+}]_0 \). After diluting the solution 10 times, the new concentration of nickel ions will be:

\[ [\text{Ni}^{2+}] = \frac{[\text{Ni}^{2+}]_0}{10} \]

### Step 3: Apply the Nernst Equation Before and After Dilution
1. **Before Dilution:**
\[ E_1 = E° - \frac{0.059}{2} \log [\text{Ni}^{2+}]_0 \]

2. **After Dilution:**
\[ E_2 = E° - \frac{0.059}{2} \log \left(\frac{[\text{Ni}^{2+}]_0}{10}\right) \]

### Step 4: Simplify the Expression for After Dilution
Using the properties of logarithms, we can rewrite the second equation:

\[ E_2 = E° - \frac{0.059}{2} \left( \log [\text{Ni}^{2+}]_0 - \log 10 \right) \]
\[ E_2 = E° - \frac{0.059}{2} \log [\text{Ni}^{2+}]_0 + \frac{0.059}{2} \cdot 1 \]
\[ E_2 = E_1 + \frac{0.059}{2} \cdot 1 \]

### Step 5: Calculate the Change in Potential
The change in potential due to dilution is:

\[ E_2 - E_1 = \frac{0.059}{2} = 0.0295 \, \text{V} \, \text{or} \, 29.5 \, \text{mV} \]

### Step 6: Determine the Final Potential
Thus, the potential of the nickel electrode after dilution increases by approximately 30 mV:

\[ E_2 = E_1 + 0.03 \, \text{V} \]

### Conclusion
The half-cell potential of the nickel electrode increases by about 30 mV after the solution is diluted 10 times.