The potential difference across the terminals of a battery of `emf 12 V` and internal resistance `2 Omega` drops to `10 V` when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is `107.9 g mol^(-1)`.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the Current Flowing Through the Circuit
We start with the formula for the electromotive force (emf) of the battery:

\[ E = I \cdot r + V \]

Where:
- \( E \) = emf of the battery = 12 V
- \( r \) = internal resistance = 2 Ω
- \( V \) = potential difference across the terminals = 10 V
- \( I \) = current (unknown)

Rearranging the equation to find the current \( I \):

\[ I = \frac{E - V}{r} \]

Substituting the known values:

\[ I = \frac{12 V - 10 V}{2 \Omega} = \frac{2 V}{2 \Omega} = 1 A \]

### Step 2: Calculate the Electrochemical Equivalent of Silver
The mass of silver deposited can be calculated using Faraday's laws of electrolysis:

\[ M = Z \cdot I \cdot T \]

Where:
- \( M \) = mass of silver deposited
- \( Z \) = electrochemical equivalent
- \( I \) = current = 1 A (from Step 1)
- \( T \) = time in seconds = 30 minutes = 1800 seconds

To find \( Z \), we use the formula:

\[ Z = \frac{E}{F} \]

Where:
- \( E \) = equivalent weight of silver = \(\frac{107.9 \text{ g/mol}}{1} = 107.9 \text{ g/mol}\) (since the valency of silver is 1)
- \( F \) = Faraday's constant = 96500 C/mol

Calculating \( Z \):

\[ Z = \frac{107.9 \text{ g/mol}}{96500 \text{ C/mol}} \approx 0.00112 \text{ g/C} \]

### Step 3: Substitute Values to Find the Mass of Silver Deposited
Now we can substitute the values into the equation for mass \( M \):

\[ M = Z \cdot I \cdot T \]

Substituting the known values:

\[ M = 0.00112 \text{ g/C} \cdot 1 \text{ A} \cdot 1800 \text{ s} \]

Calculating \( M \):

\[ M = 0.00112 \cdot 1800 \approx 2.016 \text{ g} \]

### Final Answer
The mass of silver deposited at the cathode in half an hour is approximately **2 grams**.

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