Question

$Mn{{O}_{4}}^{-}+{{C}_{2}}{{O}_{4}}^{2-}+{{H}^{+}}\to C{{O}_{2}}+{{H}_{2}}O+M{{n}^{2+}}$
The correct coefficients of $Mn{{O}_{4}}^{-},{{C}_{2}}{{O}_{4}}^{2-}$ and ${{H}^{+}}$ are respectively:
A. 2, 5, 16
B. 16, 5, 2
C. 5, 16, 2
D. 2, 16, 5

Answer 1

Hint: This question is indirectly asking us to balance this redox reaction. Hence, first we will have to find out which element is undergoing the reduction which means gain of electrons and which element is undergoing oxidation or loss of electrons, then we can separately balance them and then add them in the final step, we will get our final coefficients.

Complete Solution :
$Mn{{O}_{4}}^{-}+{{C}_{2}}{{O}_{4}}^{2-}+{{H}^{+}}\to C{{O}_{2}}+{{H}_{2}}O+M{{n}^{2+}}$
If we look at the reaction, then we look at the oxidation states of Mn which is going from +7 oxidation state to +2 which means it has gained 5 electrons and has undergone reduction and now if we look at Carbon then it is going from +3 to +4 as it has gained one electron per oxalate ion which means 2 electrons for 2 oxalate ions. Now, we know that Mn is reducing and C is getting oxidized therefore, we can write their half reactions.
Reduction Half Reaction:
$Mn{{O}_{4}}^{-}\to M{{n}^{2+}}$

- Now, we will balance this half redox reaction, since the oxidation state has changed from +7 to +2 that means it has gained 5 electrons and that should be added. Then, we see that there are 4 oxygen molecules on the left side, therefore we will add 4 ${{H}_{2}}O$ molecules on the left side to counterbalance. Then, on the right side we will add 8 hydrogen ions to balance it. The reaction will become something like this:
$8{{H}^{+}}+Mn{{O}_{4}}^{-}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$ ( 1 )

Oxidation Half Reaction:
${{C}_{2}}{{O}_{4}}^{2-}\to 2C{{O}_{2}}+2{{e}^{-}}$ (2)

- In this reaction, we have balanced the Carbon and oxygen first on the right-hand side and then the oxidation state is changed from +2 to +3 therefore, 2 electrons are lost and we have balanced them on the right-hand side.
- Now, we will add both the equations to derive our final equation after multiplying the equation (1) with 2 and equation (2) with 5 to eliminate the extra charges.
$16{{H}^{+}}+2Mn{{O}_{4}}^{-}+10{{e}^{-}}\to 2M{{n}^{2+}}+8{{H}_{2}}O$
$5{{C}_{2}}{{O}_{4}}^{2-}\to 10C{{O}_{2}}+10{{e}^{-}}$

Adding these final equations, we will get and 10 electrons will get cancelled.
$2Mn{{O}_{4}}^{-}+16{{H}^{+}}+5{{C}_{2}}{{O}_{4}}^{2-}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O$
Therefore the $Mn{{O}_{4}}^{-},{{C}_{2}}{{O}_{4}}^{2-}$ and ${{H}^{+}}$ are respectively: 2, 5 and 16.
So, the correct answer is “Option A”.

Note: There are two types of redox reactions which are carried in acidic and basic medium. This one is being carried in an acidic medium therefore, we have balanced the ions with the help of hydrogen ions, if this was carried in a basic medium then we will balance the ions with the help of hydroxide ions.