Question

$12.53$ mL of solution of $0.05093M$ $Se{O}_2$ reacted with exactly $25.52$ mL of $1MCrS{O}_4$. After the completion of the reaction what is the change of oxidation state of $Se$?

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 4

In this reaction, chromium acts as reducing agent.

${Cr}^{+2}\to {Cr}^{+3}+e^{-}$

From above reaction, we see that the change in oxidation state of chromium is 1.

Let the change is the oxidation state of Se be $a$.

$N_1{V}_1=N_2{V}_2$

$(0.05093\times a)\times 12.53=(1\times 1)\times 25.52$

$a=4$

Hence, change in the oxidation state of $Se$ is $+4$.

Therefore, option A is correct.