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0.02 equivalent of Ag was deposited in an electrolysis experiment. If same quantity of a electricity is passed through a gold solution, 1.314 g of gold is deposited. Find oxidation state of the gold. (Atomic mass of Au = 197)

Correct answer

3

Step-by-step solution

A g + + e - → Ag 1 mol 1 e q . 0.02 q. of Ag = 0.02 F = 0.02 × 96500 = 1930 C 1.314 Au is deposited by 0.02 F 197 is deposited by 0.02 1.314 × 197 = 3 F

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