Question

A current of $\text{5} \text{.0} \, \text{A}$ flows for $\text{4} \text{.0} \, \text{h}$ through an electrolytic cell containing a molten salt of metal M. This results in deposition of $\text{0} \text{.25} \, \text{mol}$ of the metal $\text{M}$ at the cathode. The oxidation state of $\text{M}$ in the molten salt is $\text{+} \, \text{x}$ . the value of ' $\text{x}$ ' is ( $\text{1} \, \text{Faraday} \, \text{=} \, \text{96000} \, \text{C}$ $mol^{- 1}$ )

Answer 1

$\text{I} = 5.0 \text{ A} , \text{ t} = 4.0 \text{ h}$

Moles of metal deposited $=\frac{1}{4}$

$Q=5.0\times 4\times 60\times 60$ coulombs

Charge in faradays

$\text{Q} = \frac{5 \times 4 \times 60 \times 60}{96000} = 0.75 \, \text{F}$

$\frac{1}{4}$ moles of metal will be deposited by $\frac{3}{4}$ F charge required

So, for 1 mole, $\frac{3}{4}\times 4=3F$

Oxidation $=+3$