Question

Vanadium electrode is oxidised electrically. If the mass of electrode decrease by 100mg during the passage of 570C. The oxidation state of vanadium in product is..

Answer 1

The solution to the given question is as follows :-

$i$ × $t = 570 C$

$w = 100mg = 100$ × $10^-^3 g$

Now, we know that,

$w =$ $(E$ × $i$ × $t)$ ÷ $F$

⇒ $(100$ × $10^-^3)$ ÷ 51 = $570$ ÷ $(96500$ × $n)$

⇒ $n = (570$ × $51)$ ÷ $(100$ × $10^-^3$ × $96500)$ = 3

Ans) The oxidation state of Vanadium in the product is 3