Question

What current strength will be required to deposit $2.692\times {10}^{-3}kg$ of Ag in $7$ minutes from ${AgNO}_3$ solution ?

Answer 1

Molecular weight of $Ag=108g/mol$

$\Rightarrow Q=1\times 7\times 60$

$\Rightarrow Q=1\times 420$

Weight= $ZQ$

$\Rightarrow Z=\frac{M}{nF}$

$\Rightarrow Z=\frac{108}{1\times F}$

$\Rightarrow \frac{108\times 7\times 60\times 1}{96500\times 2.692\times {10}^{-3}}=\frac{45360}{259.778}=174.61$ ampere.