If `Zn^(2+)//Zn` electrode is diluted 100 times, then the charge in reduction potential is

Options

A. increase of 59mV
B. decrease of 59mV
C. increase of 25.5mV
D. decrease of 2.95V

Explanation

To solve the problem of how the reduction potential changes when the concentration of the \( \text{Zn}^{2+} \) ion is diluted 100 times, we will use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation for the half-reaction of zinc can be expressed as:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{1}{[\text{Zn}^{2+}]} \right) \]
where:
- \( E \) is the electrode potential,
- \( E^\circ \) is the standard electrode potential,
- \( n \) is the number of electrons transferred (which is 2 for zinc),
- \( [\text{Zn}^{2+}] \) is the concentration of zinc ions.

### Step 2: Determine the Change in Concentration
If the concentration of \( \text{Zn}^{2+} \) is diluted 100 times, the new concentration can be represented as:
\[ [\text{Zn}^{2+}] = \frac{C}{100} = 10^{-2} \, C \]
where \( C \) is the original concentration of \( \text{Zn}^{2+} \).

### Step 3: Substitute into the Nernst Equation
Substituting the new concentration into the Nernst equation gives:
\[ E = E^\circ - \frac{0.059}{2} \log \left( \frac{1}{10^{-2}} \right) \]

### Step 4: Simplify the Logarithm
The logarithm can be simplified:
\[ \log \left( \frac{1}{10^{-2}} \right) = \log(10^{2}) = 2 \]

### Step 5: Substitute the Logarithm Back into the Equation
Substituting this back into the equation:
\[ E = E^\circ - \frac{0.059}{2} \times 2 \]
\[ E = E^\circ - 0.059 \]

### Step 6: Calculate the Change in Potential
The change in reduction potential due to the dilution is:
\[ \Delta E = E - E^\circ = -0.059 \, \text{volts} = -59 \, \text{millivolts} \]

### Conclusion
The charge in reduction potential when the \( \text{Zn}^{2+} \) electrode is diluted 100 times is \( -59 \, \text{mV} \).

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