In the electrolysis of KI, only ${\text{I}}_{\text{2}}$  is formed at the anode by the reaction;
 ${\text{2I}}^{\text{-}}\underset{ }{\overset{ }{\to }}{\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}$
After the passage of current of 0.5 ampere for 9650 seconds,  ${\text{I}}_{\text{2}}$ formed required 40 ml of 0.1 M.${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.5H}}_{\text{2}}\text{O}$  solution in the reaction;
${\text{I}}_{\text{2}}{\text{+2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\underset{ }{\overset{ }{\to }}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{2-}}{\text{+2I}}^{\text{-}}$
What is the current efficiency?