Question

A current of $3$ amperes was passed through silver nitrate solution for $125$ seconds. The amount of silver deposited at cathode was $0.42g$. Calculate the equivalent mass of silver.

Answer 1

$Ag{NO}_3+e^{-}⟶Ag$

From Faraday's first law,

$W=\frac{E\times Q}{96500}$ ($E$ is equivalent mass)

$0.42=\frac{E\times I\times t}{96500}$

$E=\frac{0.42\times 96500}{3\times 125}$

$E=108.08g$