Question

A constant current was flown for 2 hour through a $KI$ solution oxidising iodide to iodine $(2I^{-}\to I_2+2e)$. At the end of experiment liberated iodine consumed 21.75 ml of 0.0831 M solution of sodium thiosulphate following the redox change:
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
What was the average rate of current flown in ampere?

• A. $1.3A$
• B. $1.7A$
• C. $0.7A$
• D. None of these

Answer 1

The correct option is D None of these

n-factor for $S_2{O}_3^{2-}=2$

n-factor for $I_2=2$

equivalents of $I_2$=equivalents of $S_2{O}_3^{2-}$

moles of $I_2=\frac{2\times 0.0831\times 21.75\times {10}^{-3}}{2}=1.8moles$

96500 Coulomb liberates $\frac{254}{2}gm$ of $I_2$.i.e $0.5moles$

Charge required to liberate 1.8 moles$=\frac{96500\times 1.8}{0.5}=347400coulombs$

Average rate of current flown$=\frac{347400}{2\times 60\times 60}=48.25A$