A current of 1.5 A is passed through 500 mL of 0.25 M solution of zinc sulphate for 1 hr with a current efficiency of 90% . Calculate the final molarity of Zn^(2+) assuming volume to be constant .

A current of 1.5 A is passed through 500 mL of 0.25 M solution of zinc

1.	A current of 1.5 A is passed through 500 mL of 0.25 M solution of zinc sulphate for 1 hr with a current efficiency of 90% . Calculate the final molarity of Zn^(2+) assuming volume to be constant .
Answer» Solution :Quantity of ELECTRICITY passed , `Q = I xx t` `I = 1.5 xx (90)/(100) , t = 1 xx 60 xx 60 = 3600` s ` Q = 1.5 xx (90)/(100) xx 3600` `= 4860 C` Faraday of electricity passed = `(4860)/(96500) = 0.05F` Now `Zn^(2+) + 2 e^(-) to Zn` 2 F of electricity convert 1 mol of `Zn^(2+)` to Zn . 0.05 F of electricity will convert `Zn^(2+)` to Zn = `1/2 xx 0.05` `= 0.025` mol Initial moles of `Zn^(2+)` in 500 mL solution = `(0.25)/(1000) xx 500` `= 0.125` mol Moles of `Zn^(2+)` ions left in the solution = `0.125 - 0.025` = 0.10mol `therefore` Final molarity of `Zn^(2+)` solution = `(0.1)(500) xx 1000` `= 0.2 ` M