A dilute aqueous solution of `KCI `was placed between two electrodes 10 cm apart, across which a potential of 6 volt was applied. How far would the `K^(+)` ion move in 2 hours at `25^(@)C`? Ionic conductance of `K^(+)` ion at infinite dilution at `25^(@)C` is `73.52 ohm^(-1) cm^(2) "mole"^(-1)`?

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the information provided.

### Step 1: Understand the given data
- Distance between electrodes (d) = 10 cm
- Potential difference (V) = 6 V
- Ionic conductance of \( K^+ \) ion at infinite dilution (\( \lambda_m \)) = 73.52 ohm\(^{-1}\) cm\(^2\) mole\(^{-1}\)
- Time (t) = 2 hours = 2 × 60 × 60 seconds = 7200 seconds

### Step 2: Calculate the ionic mobility (\( \mu \))
The ionic mobility can be calculated using the formula:
\[ \mu = \frac{\lambda_m}{F} \]
where \( F \) (Faraday's constant) = 96500 C/mol.

Substituting the values:
\[ \mu = \frac{73.52}{96500} \approx 0.000761 \text{ cm}^2/\text{V·s} \]

### Step 3: Calculate the potential gradient (\( \nabla V \))
The potential gradient is given by:
\[ \nabla V = \frac{V}{d} \]
Substituting the values:
\[ \nabla V = \frac{6 \text{ V}}{10 \text{ cm}} = 0.6 \text{ V/cm} \]

### Step 4: Calculate the ionic speed (\( v \))
Using the relationship between ionic mobility, speed, and potential gradient:
\[ \mu = \frac{v}{\nabla V} \]
Rearranging gives:
\[ v = \mu \cdot \nabla V \]
Substituting the values:
\[ v = 0.000761 \text{ cm}^2/\text{V·s} \times 0.6 \text{ V/cm} \approx 0.0004566 \text{ cm/s} \]

### Step 5: Calculate the distance moved by \( K^+ \) ion in 2 hours
Using the formula:
\[ \text{Distance} = v \times t \]
Substituting the values:
\[ \text{Distance} = 0.0004566 \text{ cm/s} \times 7200 \text{ s} \approx 3.29 \text{ cm} \]

### Conclusion
The distance moved by the \( K^+ \) ion in 2 hours is approximately **3.29 cm**.

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