A constant current flowed for 2 hours through a potassium iodide solution oxidising the iodide ion to iodine `(2I^(-) rarr I_(2) + 2e^(-))`.
At the end of the experiment, the iodine was titrated with 21.75 mL of 0.0831 M soldium thiosulphate solution.
`(I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) +S_(4) O_(6)^(2-))`
What was the average rate of current flow in apmeres ?

Explanation

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`