When potassium permanganate, `KMnO_(4)` , is added to an acidified solution of oxalic acid, `H_(2)C_(2)O_(4)` , the products are `CO_(2)` gas and `Mn^(2+)` ions. What is the reducing agent in this reaction?

Explanation

(i) step 1 wirte the skeleton equaton for the given reaction ltbnrgt `MnO_(4)^(-)(Aq)+SO_(2)(g)rarrMn^(2+)(aq)+HSO_(4)^(-)(aq)`
step 2 Find out the element which undergo a change in O.N

here O.N of Mn decreases form +7 in `MnO_(4)^(-) "to" +2 Mn^(2+)` therefore `MnO_(4)^(-)` undergoes reductin further O.N of `SO_(2)` increases from +4 in `SO_(2)` to + 6 `HSO_(4)^(-)` therefore `SO_(2)` undergoes oxidation
step 3 divide the given skeleton Eq (a) into two half equation
reductin half equatin `MnO_(4)^(-)(aq)rarrMn^(2+)(aq)`
oxidation of half equation `SO_(2)(g)rarrHSO_(4)^(-)(aq)`
step 4 To balance reduction half eq (b) since Mn atoms on either side of Eq (b) are already balanced balance O.N by adding electrons The O.N of `MnO_(4)^(-)` is +7 while that in `Mn^(2+)` is +2 therefore each Mn atom loses 5 electrons and hence and 5 `E^(-)` to l.h.s of eq (b) we have
`MnO_(4)^(-) (aq)rarrMn^(2+)(aq)`
since the reaction occurs in acidic medium therfore balance charge by adding `H^(+)` ions ltbnrgt now total charge on the L.H.S of Eq(d) is -6 while on the R.H.S it is +2 therefore add 8 `H^(+)` to the L.H.S of Eq (d) we have
step 5 to balance oxidation half eq (c )
since S atoms on either side of equation (c ) are already balanced balance O.N by adding electrons since on of s on L.H.S of Eq (c ) is +4 while on the R.H.S it is +6 therefore and `2E^(-)` to the R.H.S of Eq (c ) we have
`SO_(2)rarrHSO_(4)^(-)(aq)+2e^(-)`
Now balance charge by adding `H^(+)` ions since the total charge on R.H.S of Eq (s) is -3 and zero on L.H.S and 3 `H^(+)` to R.H.S of Eq (g) we have `SO_(2)(g)rarrHSO_(4)^(-)(aq)+3H^(+)(Aq)+2e^(-)`
Now balance O atoms by adding `H_(2)O` molecules since there are two o atoms on the L.H.S and four or the R.H.S of Eq (h) add add `2H_(2)O` to the L.H.S of Eq (h) we have
`SO_(2)(aq)+2H_(2)O(l) rarrHSO_(4)^(-)(aq)+3H^(+)(aq)+2e^(-)`
The number of H atoms are automaticaly balanced thus eq (i) repersents the balance oxidation half equation
step 6 To cancel the number of electrons gained in reduction half equ (f) and numbr of electrons lost in the oxidation half eq (i) multilply eq (f) by 2 and eq (i) by 5 and add them together we have
`l 2MnO_(4)^(-)(aq)+16H^(+)(aq)+10e^(-)rarr2Mn^(2+)(Aq)+8H_(2O(l)`
`5SO_(2)(g)+10H_(2)O(l)rarr5HSO_(4)^(-) (aq)+15 H^(+)(aq)+10E^(-)`
`2MnO_(4)^(-)(aq)+5SO_(2)(g)+2H_(2)O(l)+H^(+)(aq)rarr2Mn^(2-)(aq)+5HSO_(4)^(-)(aq)` This repersent the final balanced redox equation
(iii) step 1 write out the element which undergo a change in O.N

here O.N of `CI_(2)` decreases from +7 in `CI_(2)O_(7)` to +3 in `CI_(2)^(-)` therefore reduce further on of o increases from -1 `H_(2)O_(2)` to 0 `O_(2)` (g) therefore `H_(2)O_(2)` undergoes oxidation