Question

Find the amount of silver liberated at the cathode if 0.500 A of current is passed through an AgNO3 electrolyte for 1 hour. Atomic weight of silver is 107.9 g mol−1.

Answer 1

The amount of silver liberated is 2.01 grams.

Explanation:

Equivalent mass of silver, $\mathrm{E}_{\mathrm{Ag}}=107.9 \mathrm{g}$........ (Ag is monoatomic)

The ECE of silver,

$Z_{A g}=\frac{E_{A g}}{f}=\frac{107.9}{96500}=0.001118$

When the formula is used, m = Zit, we obtain:  

m = 0.500 × 0.00118 × 3600 = 2.01 g  

So, the amount of silver liberated is 2.01 grams.