82. In a zinc-copper voltaic cell, if the mass of the zinc electrode decreases by 1.30 grams over time, how many coulombs of charge have passed through the circuit? (Given: Molar mass of Zn = 65.38 g/mol, Faraday's constant = 96,485 C/mol)
Key Concept: Redox reactions, Applications of voltaic cells
b) 3,839 C
[Solution Description]
First, calculate the moles of zinc oxidized:
$\text{Moles of Zn} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.30\,\text{g}}{65.38\,\text{g/mol}} \approx 0.0199\,\text{mol}$
The oxidation reaction is:
$\text{Zn (s)} \rightarrow \text{Zn}^{2+} + 2e^-$
From stoichiometry, 1 mole of Zn produces 2 moles of electrons. So, total charge $Q$ is:
$Q = n \times F \times \text{moles of electrons per mole} = 0.0199\,\text{mol} \times 96,485\,\text{C/mol} \times 2 \approx 3,839\,\text{C}$
Your Answer is correct.
b) 3,839 C
[Solution Description]
First, calculate the moles of zinc oxidized:
$\text{Moles of Zn} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.30\,\text{g}}{65.38\,\text{g/mol}} \approx 0.0199\,\text{mol}$
The oxidation reaction is:
$\text{Zn (s)} \rightarrow \text{Zn}^{2+} + 2e^-$
From stoichiometry, 1 mole of Zn produces 2 moles of electrons. So, total charge $Q$ is:
$Q = n \times F \times \text{moles of electrons per mole} = 0.0199\,\text{mol} \times 96,485\,\text{C/mol} \times 2 \approx 3,839\,\text{C}$