Source: High school physics (Chinese)
Problem
The molar mass of copper is $\mu = 63.6$ g/mol and its valence is $n = 2$. A mass of $m = 191$ g of copper is to be deposited from a copper sulfate ($\text{CuSO}_4$) solution by electrolysis. Take Faraday's constant $F = 9.65 \times 10^4$ C/mol, the elementary charge $e = 1.6 \times 10^{-19}$ C, and Avogadro's number $N_A = 6.02 \times 10^{23}$ /mol.
- How much electric charge is required?
- If the deposition is completed over one day and night (24 hours), how large must the current be?
- How many copper ions arrive at the electrode?
- What is the electrochemical equivalent $k$ of copper?
- Under a voltage of $U = 12$ V, how much electrical energy must be supplied?
(1) $Q \approx 5.80 \times 10^5$ C; (2) $I \approx 6.71$ A; (3) $N \approx 1.81 \times 10^{24}$ ions; (4) $k \approx 3.30 \times 10^{-4}$ g/C; (5) $W \approx 6.96 \times 10^6$ J
Faraday's law of electrolysis: $m = \dfrac{\mu}{n F} Q$, so $Q = \dfrac{m n F}{\mu}$.
\textbf{(1)} Charge required:
$$Q = \frac{m n F}{\mu} = \frac{191 \times 2 \times 9.65 \times 10^4}{63.6} \approx 5.80 \times 10^5\ \text{C}.$$\textbf{(2)} Time $t = 24 \times 3600 = 8.64 \times 10^4$ s. Current:
$$I = \frac{Q}{t} = \frac{5.80 \times 10^5}{8.64 \times 10^4} \approx 6.71\ \text{A}.$$\textbf{(3)} Each $\text{Cu}^{2+}$ ion carries charge $n e$, so the number of ions deposited is
$$N = \frac{Q}{n e} = \frac{5.80 \times 10^5}{2 \times 1.6 \times 10^{-19}} \approx 1.81 \times 10^{24}.$$(Equivalently, $N = \dfrac{m}{\mu} N_A = \dfrac{191}{63.6} \times 6.02 \times 10^{23} \approx 1.81 \times 10^{24}$.)
\textbf{(4)} Electrochemical equivalent (mass deposited per unit charge):
$$k = \frac{m}{Q} = \frac{\mu}{n F} = \frac{63.6}{2 \times 9.65 \times 10^4} \approx 3.30 \times 10^{-4}\ \text{g/C}.$$\textbf{(5)} Electrical energy supplied:
$$W = Q U = 5.80 \times 10^5 \times 12 \approx 6.96 \times 10^6\ \text{J}.$$