Question

Calculate the time needed for a constant current of 0.961 a to deposit 0.500 g of co(ii) as

Answer 1

1.70 × 10³ seconds

Explanation

$\text{Co}^{2+}$ + 2 e⁻ → $\text{Co}$

It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.

Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains $0.500 / 58.933 = 8.484\times 10^{-3} \; \text{mol}$ of Co atoms. It would take $2 \times 8.484 \times 10^{-3} = 0.01697 \; \text{mol}$ of electrons to reduce cobalt (II) ions and produce the $8.484\times 10^{-3} \; \text{mol}$ of cobalt atoms.

Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of $1.637 \times 10^{3} \; \text{C}$. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take $1.637 \times 10^{3} / 0.961 = 1.70 \times 10^{3} \; \text{s}$ to transfer all these charge and deposit 0.500 g of Co.