Question

In the following reaction,
$xM{n}_2{O}_7⟶yMn{O}_2+z{O}_2$ x, y, and z respectively are:

• A. 2, 4, 3
• B. 2,3,4
• C. 4,3,2
• D. None of these

Answer 1

The correct option is A

2, 4, 3

$M{n}_2{O}_7$ is splitted into two parts ( $\stackrel{+7\times 2}{M}{n}_2$ and ${\stackrel{-2\times 7}{O}}_7$ )

Let the following equation be numbered ...(i)

(i). $6e^{-}+M{n}_2\to 2Mn$
$2x=14$ $2x=8$
so change in oxidation state require $=14-8=6e^{-}$

Let the following equation be numbered ...(ii)

(ii) $2O_7\to 7O_2+28e^{-}$
$14x=-28$ $14x=0$
so change in oxidation state require $0-(-28)=28e^{-}$

Now, multiply equation (i) by 28 and equation (ii) by 6 and add them as

$28\times 6e^{-}+28M{n}_2^{7+}\to 56M{n}^{4+}$
$12O_7^{2-}\to 42O_2+28\times 6e^{-}$
overall reaction: $28M{n}_2^{7+}+12O_7^{2-}\to 56M{n}^{4+}+42O_2$
So,
$28M{n}_2{O}_7+12M{n}_2{O}_7⟶56Mn{O}_2+42O_2$ ...(iii)

In the above equation (iii), balance the $Mn$ atoms disregarding $O$ atoms.

$40M{n}_2{O}_7⟶56Mn{O}_2+24Mn{O}_2+42O_2$ ...(iv)

In the above equation, now balance the $O$ atoms to get

$40M{n}_2{O}_7⟶80Mn{O}_2+42O_2+18O_2$

The final, balanced equation is:

$2M{n}_2{O}_7⟶4Mn{O}_2+3O_2$