Question

A decinormal solution of $AgN{O}_3$ was electrolysed between Pt electrodes. After passing a small current for two hours, a fall of concentration of 0.0005124 equivalent occurred in anodic solution. The mass of copper deposited in copper coulometer placed in series was found to be 0.03879 g. Find the transport number of $N{O}_3^{-}$ in $AgN{O}_3$. Write answer to the nearest integer after multiplying transport number calculated with 10.

Answer 1

Fall of concentration around anode $=0.0005124$ equivalent
Mass of copper deposited in coulometer $=0.03879g$
$\therefore$ Equivalent of copper deposited in coulometer
$=\frac{0.03879}{31.9}=0.001216equivalent$
Now transport no. $t_{A{g}^{+}}$
$=\frac{\text{No. of equivalent of}A{g}^{+}\text{ion lost in anode}}{\text{No. of equivalent of}C{u}^{2+}\text{deposited in coulometer}}$
or total fall in concentration of solution
$=\frac{0.0005124}{0.001216}$
$t_{A{g}^{+}}=0.4214$
Hence, the transport no. of $A{g}^{+}$ is 0.4214.
Also, $t_{A{g}^{+}}+t_{N{O}_3^{-}}=1$
$t_{N{O}_3^{-}}=1-0.4214=0.5786$
Thus, the transport no. of $N{O}_3^{-}$ is $0.5786\approx 6$.