Question

Two voltameters containing CuSO₄ and acidulated water respectively are connected in series and the same current is passed for some time. 0.3177 g copper is deposited in the first one. Calculate the weight of liberated hydrogen in the second. (At. wt. of Cu = 63.5 and H = 1.008)

Answer 1

\[\ce{Cu^2+ + 2e- -> Cu}\]

2 Faradays deposit 1 mole (63.5 g) of Cu.

So, the charge required per gram:

1 g Cu = `2/63.5` mol e⁻

Thus, for 0.3177 g Cu:

mol e⁻ = `(2 xx 0.3177)/63.5`

= 0.010 mol e⁻

Reaction in water volameter:

\[\ce{2H+ + 2e− -> H2}\]

2 Faradays liberate 1 mole (2.016 g) of H₂.

So, 1 Faraday liberates 1.008 g of H

For 0.010 mol e⁻:

Mass of H = 0.010 × 1.008 

= 0.01008 g