Question

When $9.65$ coulomb of electricity is passed through a solution of silver nitrate, the amount of silver deposited is:

• A. $16.2mg$
• B. $21.2mg$
• C. $10.8mg$
• D. $6.4mg$

Answer 1

Answer: (C)

$10.8mg$

According to Faraday's Law of Electrolysis:-

$W=\frac{I\times t\times M}{nF}$

$W$= weight deposited

$I$= current $F$= Faraday's constant $96500$

$t$= time $M$= Molar mass

$n$= number of electron

Given:-

$It=9.65$ coulomb

Molar mass of Silver, $M=108g/mol$ $n=1$

$W=\frac{9.65\times 108}{1\times 96500}$

$W=0.0108g$

$\Rightarrow W=10.8mg$