Question

$xKMn{O}_4+y{K}_2{C}_2{O}_4+z{H}_{2}S{O}_4⟶aMn{O}_4+bC{O}_2+6K_{2}S{O}_4+8H_{2}O$

Here x,y,z, a and b are respectively

• A. 2, 5, 8, 2, 10
• B. 2, 4, 9, 4, 10
• C. 5, 9, 12, 4, 10
• D. None of these

Answer 1

The correct option is A

2, 5, 8, 2, 10

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

$Mn{O}_4^{-}$ + $C_2{O}_4^{-2}$ $⟶$$M{n}^{2+}$ + $C{O}_2$

Note that O and H atoms attached to the central atom 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 that has undergone the change in oxidation state is balanced.

b.Balance O by adding $4H_{2}O$ to the 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 is1$\times$(-1) + 8$\times$(+1) = +7

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

Add $5e^{-}$ in LHS ( each $e^{-}$ is equivalent to a charge of -1)

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

Oxidation:

$C_2{O}_4^{-2}$$⟶$$2C{O}_2$

Following the same procedure as above, we have:

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

b. Balance O atoms :Already balanced

c. Balanced H atoms: There are none

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

MultiplyingEq.(i )by 5 to balance the electrons transfer add to get

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

e.Now add other ions to both sides

L.H.S $12K^{+}$ ; $8S{O}_4^{2-}$
R.H.S $12K^{+}$ ; $8S{O}_4^{2-}$

f. $2K^{+}+2Mn{O}_4^{-}+10K^{+}+5C_2{O}_4^{2-}+8S{O}_4^{2-}+16H^{+}⟶$
$2M{n}^{+2}+2S{O}_4^{-2}+10C{O}_2+12K^{+}+6S{O}_4^{-2}+8H_{2}O$

$2KMn{O}_4+5K_2{C}_2{O}_4+8H_{2}S{O}_4⟶2MnS{O}_4+10C{O}_2+6K_{2}S{O}_4+8H_{2}O$