Question

Consider the following reaction

$xMn{O}_4^{-}+y{C}_2{O}_4^{2-}+z{H}^{+}⟶M{n}^{+2}+2yC{O}_2+\frac{z}{2}{H}_{2}O$

the values of x,y and z in the reaction are, respectively:

• A. 2,5 and 16
• B. 5,2 and 8
• C. 5,2 and 16
• D. 2,5 and 8

Answer 1

The correct option is A

2,5 and 16

Ion electron method: First write the given equation in ionic form having ions with central atom (which has undergone a change in oxidation state).

$xMn{O}_4^{-}+y{C}_2{O}_4^{2-}+z{H}^{+}⟶M{n}^{+2}+2yC{O}_2+\frac{z}{2}{H}_{2}O$

Note that O and H atoms attached to the central atom (shown in underlined) have to be retained.

Now, write the oxidation and reduction half reaction and balance them as shown:

Reduction:

$Mn{O}_4^{-}+C_2{O}_4^{-2}⟶M{n}^{+2}$

a. First , make sure that the element undergone the change in oxidation state is balanced.

b. Balance O atoms by adding $4H_2$ on RHS.

$Mn{O}_4^{-}⟶M{n}^{+2}+4H_{2}O$

c.Now, RHS has excess of 8H atoms. Add $8H^{+}$ on LHS . Note, the medium is acidic due to the presence of $H_{2}S{O}_4$ .

$2Mn{O}_4^{-}+8H^{+}⟶2M{n}^{+2}+4H_{2}O$

d. Now O and H are balanced . Balance the charge on both sides.

On LHS: Charge is 1$\times$ + 8$\times$(+1) = +7

On RHS: Charge is 1$\times$2 + 4$\times$(0) = +2

Add 5$e^{-}$ in LHS (Note : each $e^{-}$ is equivqlent to acharge of -1)

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{+2}+4H_{2}O$ ...(i)

Oxidation:

$C_2{O}_4^{-2}⟶C{O}_2$

Following the same procedure as above, we have:

a. Balance C atoms :$C_2{O}_4^{-2}⟶C{O}_2$

b.Balance O atoms : Already balanced

c.Balance H atoms: No H atoms

d.$C_2{O}_4^{-2}⟶2C{O}_2+2e^{-}$ ...(ii)

Multiply Eq(i)by 2 and Eq.(ii) by5 to balance the electrons transfer and add to get

$2Mn{O}_4^{-}+5C_2{O}_4^{-2}+16H^{+}⟶2M{n}^{+2}+10C{O}_2+8H_{2}O$