Question

Permanganate solution reacts with the oxalate ion in acidic medium and 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. 1
• C. 3
• D. 5

Answer 1

The correct option is A 2

The redox reaction is :
$Mn{O}_4^{-}+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$
$\stackrel{+7}{M}n{O}_4^{-}+{\stackrel{+3}{C}}_2{O}_4^{2-}\to \stackrel{+2}{M}{n}^{2+}+\stackrel{+4}{C}{O}_2$
$Mn{O}_4^{-}$ is oxidising agent
$C_2{O}_4^{2-}$ is reducing agent.

$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|)\times \text{number of atom}$
${\stackrel{+3}{C}}_2{O}_4^{2-}\to \stackrel{+4}{C}{O}_2$---oxidation
$n_f=(|4-3|\times 2)=2$
$\stackrel{+7}{M}n{O}_4^{-}\to \stackrel{+2}{M}{n}^{2+}$----reduction
$n_f=(|2-7|\times 1)=5$
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.