A current of $$10.0A$$  flows for $$2.00\,h$$  through an electrolytic cell containing a molten salt of metal $$X.$$ This results in the decomposition of $$0.250\,mol$$   of metal $$X$$ at the cathode. The oxidation state of $$X$$ in the molten salt is : $$\left( {F = 96,500\,C} \right)$$

Solution :
According to Faraday’s first law of electrolysis
$$\eqalign{ & W = \frac{{E \times i \times t}}{{96500}} \cr & {\text{Where }}E{\text{ = equivalent weight}} \cr & = \frac{{mol.{\text{ }}mass{\text{ of metal}}(M)}}{{{\text{oxidation state of metal }}(x)}} \cr & {\text{Substituting the value in the formula}} \cr & W = \frac{M}{x} \times \frac{{i \times t}}{{96500}} \cr & {\text{or}}\,\,x = \frac{M}{W} \times \frac{{i \times t}}{{96500}} \cr & = \frac{{10 \times 2 \times 60 \times 60}}{{96500 \times 0.250}} \cr & = 3 \cr & \left[ {{\text{Given : no}}{\text{. of }}moles = \frac{M}{W} = 0.250} \right] \cr & {\text{Hence oxidation state of metal is}}\left( { + 3} \right) \cr} $$