Question

Answer the following:

What current strength in amperes will be required to produce 2.4 g of Cu from CuSO₄ solution in 1 hour? Molar mass of Cu = 63.5 g mol^–1.

Answer 1

Given:

Mass of Cu = 2.4 g,
Molar mass of Cu = 63.5 g mol^–1
1 hours = 1 × 60 × 60s = 3600s

To find: Current strength (in amperes)

Formulae: 

1) Mole ratio = ${\displaystyle \frac{\text{Moles of product formed in half reaction}}{\text{Moles of electrons required in half reaction}}}$

2) W = ${\displaystyle \frac{\text{I}\left(\text{A}\right)\times \text{t}\left(\text{s}\right)}{96500(\text{C}/\text{mol} {\text{e}}^{-})}\times \text{mole ratio}\times \text{molar mass}}$

Calculation:

1) Stoichiometry for the formation of Cu is

${\displaystyle {\text{Cu}}_{\text{s}}^{2+}+2{\text{e}}^{-}\to {\text{Cu}}_{\left(\text{s}\right)}}$

Using formula (i),

Mole ratio = ${\displaystyle \frac{1 \text{mole}}{\text{2 mole}}}$

2) Using formula (ii),

W = ${\displaystyle \frac{\text{I}\left(\text{A}\right)\times \text{t}\left(\text{s}\right)}{96500(\text{C}/\text{mol} {\text{e}}^{-})}\times \text{mole ratio}\times \text{molar mass}}$

${\displaystyle 2.4\text{g}=\frac{\text{I}\left(\text{A}\right)\times \text{t}\left(\text{s}\right)}{96500(\text{C}/\text{mol} {\text{e}}^{-})}\times \frac{1 \text{mole}}{\text{2 mole} {\text{e}}^{-1}}\times 63.5{\text{g mol}}^{-1}}$

I(A) = ${\displaystyle \frac{2.4\times 96500\times 2}{63.5\times 3600}}$ = 2.03 A

Current strength in amperes required to produce 2.4 g of Cu from CuSO₄ is 2.03 A.