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 these calculations:

### Step 1: Calculate the Cell Constant (G*)
The cell constant (G*) is given by the formula:
\[ G^* = \frac{L}{A} \]
Where:
- \( L \) = distance between the electrodes = 0.50 cm
- \( A \) = surface area of the electrodes = 1.50 cm²

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

### Step 2: Calculate the Volume Containing 1 Equivalent
The normality (N) of the solution is given as 0.10 N. This means:
- 0.10 equivalents are present in 1 liter (1000 mL).

To find the volume (V) containing 1 equivalent:
\[ V = \frac{1000 \, \text{mL}}{0.10} = 10000 \, \text{mL} \]

### Step 3: Calculate the Conductivity (κ)
The equivalent conductance (Λ) is given as 97.1 mho cm²/equivalent. The conductivity (κ) can be calculated using the formula:
\[ \kappa = \frac{\Lambda}{V} \]
Where:
- \( \Lambda = 97.1 \, \text{mho cm}^2/\text{eq} \)
- \( V = 10000 \, \text{mL} \)

Substituting the values:
\[ \kappa = \frac{97.1}{10000} = 9.71 \times 10^{-3} \, \text{mho cm}^{-1} \]

### Step 4: Calculate the Resistance (R)
Using the relationship between conductivity and resistance:
\[ R = \frac{G^*}{\kappa} \]
Substituting the values:
\[ R = \frac{1/3}{9.71 \times 10^{-3}} \]

Calculating this gives:
\[ R = \frac{0.3333}{0.00971} \approx 34.33 \, \Omega \]

### Step 5: Calculate the Current (I) using Ohm's Law
Ohm's Law states:
\[ V = R \cdot I \]
Rearranging gives:
\[ I = \frac{V}{R} \]
Where:
- \( V = 5 \, \text{V} \)
- \( R = 34.33 \, \Omega \)

Substituting the values:
\[ I = \frac{5}{34.33} \approx 0.1456 \, \text{A} \]

### Final Answer
The current that will flow when the potential difference between the electrodes is 5 volts is approximately **0.1456 A**.

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