Question

A current of 2 A is passed for $1.93\times {10}^4$ s through a molten tin salt depositing 23.8 g Sn(A.W. of Sn = 119). The oxidation state of Sn in the salt is:

Answer 1

$I=2At=1.93\times {10}^{4}sW=23.8gSnM=119g{mol}^{-1}n=\frac{23.8}{119}W=ZItQ=n\times n_f\times F2\times 1.93\times {10}^4=\frac{23.8}{119}\times n_f\times 96500\frac{2\times 1.93\times {10}^4\times 119}{23.8\times 96500}=n_{f}2\times {10}^{-4}\times {10}^4=n_f{n}_f=2$

Oxidation state = $2$