A constant current was flowing for 2 hours through a KI solution oxidizing iodide ions to iodine (2I^(-)toI_(2)+2e^(-)). At the end of the experiment, liberated iodine consumed 21.75 mL of 0.0831 M sodium thisulphate solution following the redox change I_(2)+2S_(2)O_(3)^(2-)to2I^(-)+S_(4)O_(6)^(2-). calculate the average rate of current flow.

A constant current was flowing for 2 hours through a KI solution oxidi

1.	A constant current was flowing for 2 hours through a KI solution oxidizing iodide ions to iodine (2I^(-)toI_(2)+2e^(-)). At the end of the experiment, liberated iodine consumed 21.75 mL of 0.0831 M sodium thisulphate solution following the redox change I_(2)+2S_(2)O_(3)^(2-)to2I^(-)+S_(4)O_(6)^(2-). calculate the average rate of current flow.
Answer» SOLUTION :21.75 mL of 0.831 M `Na_(2)S_(2)O_(3)=(0.0831)/(1000)xx21.75`mole as 2 moles of `Na_(2)S_(2)O_(3)` react with 1 mole of `I_(2)` `thereforeI_(2)` liberated`=(1)/(2)xx(0.0831)/(1000)xx21.75`mole 1 mole of `I_(2)` is liberated by 2F or `2xx96500`COULOMBS `therefore(0.0831xx21.75)/(2000)` mole will be liberated by CHARGE `=2xx96500xx(0.0831)/(2000)`coulombs=174.4 coulombs `Q=Ixxt` or `I=(Q)/(t)=(174.4)/(2xx60xx60)=0.0242A`