A constant current was flown for one minute through a solution of KI. At the end of experiment, liberated I₂ consumed 150ml of 0.01Msolution of Na₂S₂O₃. Average rate of current flow is

When electricity passed throught the KI solution I₂ is llibetated ,The I₂ liberated is reacted with Na₂S₂O₃. The reactions taking place are , 

$2S_2{O}_3^{2-}\to S_4 O_6^{2-}+ 2e^{-} (\hspace{0.17em}n- factor= \frac{2}{2}=1)$

$$\begin{gathered} \mathrm{milli}. \mathrm{Eq}. \mathrm{of} \mathrm{KI}\hspace{0.17em}= \mathrm{milli}. \mathrm{Eq} \mathrm{of} {\mathrm{I}}_2 \mathrm{liberated} \\ = \mathrm{milli}. \mathrm{Eq} \mathrm{of} {\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3 \\ = 150 \times 0.01\times 1(\mathrm{n}-\mathrm{factor}) \\ = 1.5 \mathrm{milli}.\mathrm{Eq} \\ = 1.5 \times {10}^{-3 } \mathrm{Eq}. \\ \mathrm{since} 96500 \mathrm{coulombs} = 1 \mathrm{Eq}. \\ 1.5 \times {10}^{-3} \mathrm{Eq}= 96500 \times 1.5\times {10}^{-3 }\mathrm{coulombs} \\ \mathrm{Q}= \mathrm{it} \\ \mathrm{i}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{96500\times 1.5\times {10}^{-3}}{60 \sec } \\ \mathrm{on} \mathrm{solving} \\ \mathrm{i}=2.4125 \mathrm{amp} \end{gathered}$$