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For the redox reaction, $MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow{{}}M{{n}^{+2}}+C{{O}_{2}}+{{H}_{2}}O$. The correct coefficient of the reactant for the balanced equation are:(A)$MnO_{4}^{-}$ = 2; ${{C}_{2}}O_{4}^{-2}$ = 16; ${{H}^{+}}$ = 5(B) $MnO_{4}^{-}$ = 16; ${{C}_{2}}O_{4}^{-2}$ = 5; ${{H}^{+}}$ = 2(C) $MnO_{4}^{-}$ = 5; ${{C}_{2}}O_{4}^{-2}$ = 16; ${{H}^{+}}$ = 2(D) $MnO_{4}^{-}$ = 2; ${{C}_{2}}O_{4}^{-2}$ = 5; ${{H}^{+}}$ = 16

Last updated date: 20th Jun 2024
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Hint: It is a redox reaction, so we can write reaction into two parts oxidation and reduction. The representation of splitting and writing the reaction is known as the half reaction. Firstly, to answer the question we have to write the half reaction and followed by balancing.

Redox reaction is a chemical reaction where both the oxidation as well as reduction reaction takes place simultaneously.
Oxidation process is a process where electrons are lost in the course of the reaction and reduction process is a process where the electrons are gained.
$MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow{{}}M{{n}^{+2}}+C{{O}_{2}}+{{H}_{2}}O$
Now, let's write the half reaction:
Oxidation: ${{C}_{2}}O_{4}^{-2}\to C{{O}_{2}}$
Reduction: $MnO_{4}^{-}\to M{{n}^{+2}}$
Now, let's balance this equation:
- Firstly, let's balance the atoms other than O and H in each half reaction. If we observe the above written half reaction, the number of carbon atoms is 2 in the reactant side but there is only one carbon in the product side (oxidation). So, we have to add 2 before $C{{O}_{2}}$ in order to balance the equation. In the second part (reduction) the number of Mn is equal in both sides.
Oxidation: ${{C}_{2}}O_{4}^{-2}\to 2C{{O}_{2}}$
Reduction: $MnO_{4}^{-}\to M{{n}^{+2}}$
- Now, let's balance the O atom in the half reaction.
In the oxidation process the O atom on both sides are equal. In the second part (reduction) there are four O atoms in reactant side but, no O atom in product so, firstly let's add an water molecule in product followed by balancing.
Oxidation: ${{C}_{2}}O_{4}^{-2}\to 2C{{O}_{2}}$
Reduction: $MnO_{4}^{-}\to M{{n}^{+2}}+4{{H}_{2}}O$
-Now, let's balance the H atom on both sides
Only the second part is unbalanced as there are 8 H atoms in the product but there is no H atom in the reactant side so, let's add ${{H}^{+}}$ followed by balancing.
Oxidation: ${{C}_{2}}O_{4}^{-2}\to 2C{{O}_{2}}$
Reduction: $MnO_{4}^{-}+8{{H}^{+}}\to M{{n}^{+2}}+4{{H}_{2}}O$
-Now, let's balance the charge
Oxidation: ${{C}_{2}}O_{4}^{-2}\to 2C{{O}_{2}}+2{{e}^{-}}$
Reduction: $MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{+2}}+4{{H}_{2}}O$
In order to balance the reaction let's multiply oxidation by 5 and reduction by 2.
Oxidation: $5{{C}_{2}}O_{4}^{-2}\to 10C{{O}_{2}}+10{{e}^{-}}$
Reduction: $2MnO_{4}^{-}+16{{H}^{+}}+10{{e}^{-}}\to 2M{{n}^{+2}}+8{{H}_{2}}O$
$5{{C}_{2}}{{O}_{4}}+2Mn{{O}_{4}}+16{{H}^{+}}\to 10C{{O}_{2}}+2M{{n}^{+2}}+8{{H}_{2}}O$
Never forget to balance the charge, this is where most of it may go wrong. In balancing O and H atoms we added ${{H}_{2}}O$ and ${{H}^{+}}$. This was done as this reaction took place in an acidic medium. The number of atoms and charges should be the same in both reactant as well as product side; this is the only criteria that we should remember.