Calculate the mass of copper deposited if a current of 0.45A flows through \(CuSO_{4}\) solution for 1hour 15mins. [Cu=64.0, S= 32.0, O =16.0, 1F= 96500C]
A. 6.40g B. 0.67g C. 0.64g D. 0.45g
Correct Answer: B
Explanation
\(Cu^{2+} + 2e^{-} \to Cu_{(s)}\) \( 2 \times 96500C = 193000C deposits 64g of Cu\) \(M \propto It\) \(Q = It = 0.45A \times 1hr15mins\) = \( 0.45 \times 75 \times 60\) [converting the time to seconds] Quantity of electricity passed = 2025C \(193000C \to 64g Cu\) \(1C \to \frac{64}{193000}\) \(2025C \to \frac{64}{193000} \times 2025\) \(= 0.6715g \approxeq 0.67g\)
