Question

The reaction of the permanganate ion with the oxalate ion in acidic solution forms manganese(II) ion and carbon dioxide. What is the stoichiometric coefficient in front of manganese(II) in the balanced redox equation?
$Mn{O}_4^{-}+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$

• A. 2
• B. 5
• C. 1
• D. 3

Answer 1

The correct option is A 2

Steps for Balancing redox reactions:

1. Identify the oxidation and reduction half reactions
2. Find the oxidising and reducing agent.
3. Find the n-factor of oxidising and reducing agent.
4. Balance atom undergoing oxidation and reduction.
5. Cross multiply the oxidising or reducing agent with simplified n-factor values
6. Balance atoms other than oxygen and hydrogen.
7. Balancing oxygen atoms
8. Balancing hydrogen atoms
9. Balance charge

For acidic medium:
As soon as we add x $H_{2}O$ units, we add 2x $H^{+}$ ions on the opposite side.

$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$
$\stackrel{+7}{M}n{O}_4^{-}+{\stackrel{+3}{C}}_2{O}_4^{2-}\to \stackrel{+2}{M}{n}^{2+}+\stackrel{+4}{C}{O}_2$

${\stackrel{+3}{C}}_2{O}_4^{2-}\to \stackrel{+4}{C}{O}_2$oxidation
$n_f=(|4-3|\times 2=2$
$\stackrel{+7}{M}n{O}_4^{-}\to \stackrel{+2}{M}{n}^{2+}$ reduction
$n_f=(|2-7|\times 1=5$
$Mn{O}_4^{-}$ is oxidising agent
$C_2{O}_4^{2-}$ is reducing agent.

Balance atom undergoing oxidation and reduction.
$Mn{O}_4^{-}+C_2{O}_4^{2-}\to M{n}^{2+}+2C{O}_2$
Cross mutiply the oxidising or reducing agent with simplified n-factor values.
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2$

Balance atoms oxygen.

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

balance hydrogen atom
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
Balance charge
charge in reactant side = +4
charge in product side = +4
so the balanced equation is $2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
the coefficient of $Mn$ is 2.