For `H^(+)` and `Na^(+)` the values of `lambda^(oo)` are `349.8` and `50.11`. Calculate the mobilities of these ions and their velecities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the mobilities of the ions

The mobility (U) of an ion can be calculated using the formula:

\[ U = \frac{\lambda^\infty}{F} \]

where:
- \(\lambda^\infty\) is the limiting molar conductivity of the ion,
- \(F\) is Faraday's constant, which is approximately \(96500 \, \text{C/mol}\).

#### For Hydrogen Ion (\(H^+\)):
Given:
\(\lambda^\infty (H^+) = 349.8 \, \text{S cm}^2/\text{mol}\)

Substituting the values:

\[ U_{H^+} = \frac{349.8}{96500} = 3.62 \times 10^{-3} \, \text{cm}^2/\text{V s} \]

#### For Sodium Ion (\(Na^+\)):
Given:
\(\lambda^\infty (Na^+) = 50.11 \, \text{S cm}^2/\text{mol}\)

Substituting the values:

\[ U_{Na^+} = \frac{50.11}{96500} = 5.20 \times 10^{-4} \, \text{cm}^2/\text{V s} \]

### Step 2: Calculate the velocities of the ions

The velocity (v) of an ion can be calculated using the formula:

\[ v = U \cdot \frac{V}{d} \]

where:
- \(U\) is the mobility of the ion,
- \(V\) is the potential difference applied,
- \(d\) is the distance between the electrodes.

Given:
- \(V = 2 \, \text{V}\)
- \(d = 5 \, \text{cm}\)

#### For Hydrogen Ion (\(H^+\)):
Substituting the values:

\[ v_{H^+} = 3.62 \times 10^{-3} \cdot \frac{2}{5} = 1.45 \times 10^{-3} \, \text{cm/s} \]

#### For Sodium Ion (\(Na^+\)):
Substituting the values:

\[ v_{Na^+} = 5.20 \times 10^{-4} \cdot \frac{2}{5} = 2.08 \times 10^{-4} \, \text{cm/s} \]

### Final Results:
- The mobility of \(H^+\) is \(3.62 \times 10^{-3} \, \text{cm}^2/\text{V s}\).
- The mobility of \(Na^+\) is \(5.20 \times 10^{-4} \, \text{cm}^2/\text{V s}\).
- The velocity of \(H^+\) is \(1.45 \times 10^{-3} \, \text{cm/s}\).
- The velocity of \(Na^+\) is \(2.08 \times 10^{-4} \, \text{cm/s}\).