Question

Calculate the number of coulombs required to deposit $50g$ of silver at cathode from silver nitrate solution.
(Atomic mass of silver $=108$).

Answer 1

$\Reduction Half Reaction:-

$A{g}^{+}+e^{-}\to A{g}_{\left(s\right)}$.

(i) mole Ratio$=\frac{molesofAgformed}{molesof{e}^{-}required}=\frac{1}{1}=1$.

(ii)Moles of $Ag$ produced$=\frac{Q\left(C\right)}{96500Cmo{l}^{-1}}\times moleratio$

$\frac{50}{108}=\frac{Q\left(C\right)}{96500Cmo{l}^{-1}}\times 1$

$\therefore Q=\frac{50\times 96500}{108}=44675.9C$

$\therefore Q=44675.9C$.