Question

One gram of copper is deposited in a copper voltameter when a current of $0.5$ A flows for $30$ minutes. Then the current required to deposit $2$ g of silver in the same time is (ece of copper $= 3.3 \times 10^{-4}$ g C$^{-1}$, ece of silver $= 1.1 \times 10^{-3}$ g C$^{-1}$)

• A. 4 A
• B. 6 A
• C. 2 A
• D. 5 A
• E. 3 A

Hint:

Mass deposited $\propto ZI$ for same time.

Answer 1

The Correct Option is B

Concept: Faraday’s First Law: \[ m = ZIt \] where $Z$ is electrochemical equivalent.

Step 1: For copper: \[ 1 = (3.3 \times 10^{-4})(0.5)t \]

Step 2: For silver: \[ 2 = (1.1 \times 10^{-3}) I t \]

Step 3: Divide both: \[ \frac{2}{1} = \frac{1.1 \times 10^{-3} \cdot I}{3.3 \times 10^{-4} \cdot 0.5} \]

Step 4: Solve: \[ 2 = \frac{1.1}{3.3} \cdot \frac{I}{0.5} \cdot 10 \] \[ 2 = \frac{1}{3} \cdot 2I \cdot 10 = \frac{20I}{3} \] \[ I = \frac{6}{1} = 6 \text{ A} \] Final Answer: $6$ A