The equivalent conductance of `0.10 N` solution of `MgCI_(2)` is 97.1 mho `cm^(2) eq^(-1)`. A cell electrodes that are `1.50 cm^(2)` in surface are and `0.50` cm apart is filled with `0.1N MgCI_(2)` solution. How much current will flow when the potential difference between the electrodes is 5 volts?

Explanation

To solve the problem step by step, we will follow the concepts of electrochemistry, specifically focusing on equivalent conductance, specific conductance, and Ohm's law.

### Step 1: Calculate the Cell Constant (C)
The cell constant (C) is given by the formula:
\[ C = \frac{d}{A} \]
where:
- \(d\) = distance between the electrodes = 0.50 cm
- \(A\) = area of the electrodes = 1.50 cm²

Substituting the values:
\[ C = \frac{0.50 \, \text{cm}}{1.50 \, \text{cm}^2} = \frac{1}{3} \, \text{cm}^{-1} \]

### Step 2: Calculate the Specific Conductance (κ)
The specific conductance (κ) can be calculated from the equivalent conductance (Λ) using the formula:
\[ \kappa = \frac{\Lambda}{V} \]
where:
- \(\Lambda\) = equivalent conductance = 97.1 mho cm² eq⁻¹
- \(V\) = volume of 1 equivalent

For a 0.10 N solution of MgCl₂, the volume of 1 equivalent is:
\[ V = \frac{1000 \, \text{cm}^3}{0.10 \, \text{eq}} = 10000 \, \text{cm}^3 \]

Now substituting the values:
\[ \kappa = \frac{97.1 \, \text{mho cm}^2 \text{eq}^{-1}}{10000 \, \text{cm}^3} = 0.00971 \, \text{mho cm}^{-1} \]

### Step 3: Calculate the Conductance (G)
The conductance (G) is related to the specific conductance and cell constant by:
\[ G = \kappa \cdot C \]
Substituting the values:
\[ G = 0.00971 \, \text{mho cm}^{-1} \cdot \frac{1}{3} \, \text{cm}^{-1} = 0.00323667 \, \text{mho} \]

### Step 4: Calculate the Resistance (R)
The resistance (R) can be calculated as:
\[ R = \frac{1}{G} \]
Substituting the value of G:
\[ R = \frac{1}{0.00323667} \approx 309.5 \, \Omega \]

### Step 5: Calculate the Current (I) using Ohm's Law
According to Ohm's law:
\[ I = \frac{V}{R} \]
where:
- \(V\) = potential difference = 5 V

Substituting the values:
\[ I = \frac{5 \, \text{V}}{309.5 \, \Omega} \approx 0.0161 \, \text{A} \]

### Final Answer
The current that will flow when the potential difference between the electrodes is 5 volts is approximately:
\[ I \approx 0.0161 \, \text{A} \]