A constant current was flowen for `1 mi n` through a solution of `Kl` . At the end of experiment, liberated `I_(2)` consumed `150mL` of `0.01M` solution of `Na_(2)S_(2)O_(3)` following the reaction `:` `I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(c-)+S_(4)O_(6)^(2-)` What was the average rate of current flow in ampere ?

A constant current was flowen for `1 mi n` through a solution of `Kl`

1.	A constant current was flowen for `1 mi n` through a solution of `Kl` . At the end of experiment, liberated `I_(2)` consumed `150mL` of `0.01M` solution of `Na_(2)S_(2)O_(3)` following the reaction `:` `I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(c-)+S_(4)O_(6)^(2-)` What was the average rate of current flow in ampere ?
Answer» `2S_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-)+2e^(-)" "(n` factor `=(2)/(2)=1)` `mEq` of `Kl=mEq` of `I_(2)` liberated `=mEq` of `Na_(2)SO_(3)` `=150 mLxx0.01xx1(n` factor `)` `=1.5mEq` `[1F=96500C=1 Eq` or `S_(2)O_(3)^(2-)]` `1.5xx10^(-3)Eq=96500Cxx1.5xx10^(-3)Eq` `implies I xx t(s)=Ixx60` `:. I=(96500Cxx1.5xx10^(-3)Eq)/(60s)=(965)/(400)=2.4125A`

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