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A current of 1.50 A was passed through an electrolytic cell containing AgNO₃ solution with inert electrodes. The weight of the silver deposited was 1.50 g. How long did the current flow? (Molar mass of Ag = 108 g mol^–1 , 1F = 96500 C mol^–1 ).

Answers (1)

$\mathrm{I = 1.50A}$ $t = ?$

For 108 g of Ag IF = 96500C

$1g = \frac{196500}{108}$

$1.5g = \frac{196500\times 1.5}{108} = 1340.278C$

$Q = It$

$t = \frac{Q}{I} = \frac{1340.278}{1.5} = 893.518 s$

Posted by

Sumit Saini

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