Question

In refining of silver by electrolytic method what will be the weight of 100 gram silver anode if 5 ampere current is passed for 2 hours? Purity of silver is 95 % by weight

Answer 1

Answer on Question #49441 – Chemistry – Inorganic Chemistry

According to Faraday law:

$$m = \frac {I \cdot t \cdot M}{F \cdot z}$$

m is the mass of the substance liberated at an electrode in grams

I is the current passed through the substance

t is time of electrolysis

F = 96485 C mol⁻¹ is the Faraday constant

M is the molar mass of the substance

z is the valency number of ions of the substance (electrons transferred per ion).

Anode process:

Ag(s) -1e → Ag⁺ (aq)

$$m = \frac {5 A \times 7200 \, s \times 108 \, g/mol}{96485 \, C/mol \times 1} = 40.296 \, g$$

If purity of silver is 95 % by weight so:

$$m_{Ag} = \frac {40.296}{0.95} = 42.417 \, g$$

Anode is dissolving, losing mass. Δm is 42.417 g, starting mas is 100 g, so resulting would be:

$$m_{anode} = 100 \, g - 42.417 \, g = 57.583 \, g$$

Answer: 57.583 g

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