Question

A polyvalent metal weighing $0.1$ g and having atomic mass $51.0$ reacted with dilute $H_{2}S{O}_4$ to give $43.9$ mL of $H_2$ at STP. The solution containing the metal in the lower oxidation state was found to require $58.8$ mL of $0.1NKMn{O}_4$ for complete oxidation. What are the valencies of metal- assuming x and y are the two oxidation states of the metal.? (Write it as the product, $xy$)

Answer 1

The redox changes are as follows.

$M⟶M^{n+}+ne$
$2H^{+}+2e⟶H_2$

Meq. of metal $=$ Meq. of $M{n}^{n+}$ $=$ Meq. of $H_2$

$=\frac{43.9}{11200}\times 1000=3.92$ $(\because 11200$ mL $H_2=1$ equivalent$)$

$\left\{\begin{array}{c}\frac{0.1}{\left(\frac{51}{n}\right)}\end{array}\right\}\times 1000=3.92$ or $n=2$

The metal from $M^{2+}$ is now oxidized by $KMn{O}_4$.

$M^{2+}⟶M^{a+}+(a-2)e$

$M{n}^{7+}+5e⟶M{n}^{2+}$

Meq. of $M^{2+}$ $=$ Meq. of $KMn{O}_4$

$\frac{0.1}{\left(\frac{51}{a-2}\right)}\times 1000=58.8\times 0.1$ $⟹:a=5$

Thus, different oxidation states of metal are $2$ and $5$.

So, the answer is $2\times 5=10$