In passing chlorine gas through a concentrated solution of alkali we get chloride and chlorate ions Obtain balanced chemical equation for this reaction.

Explanation

Step 1 Write the skeletal eqation for the given reaction
`CI_(2)(g)+OH^(-)rarrCI^(-)(aq)+CIO_(3)^(-)(aq)`
Step 2 Write the O.N of al the elements above their respective sysmbols

Total increase `=2xx5=10`
Total decrease `=2 xx-1=-2`
Step 3 Find out hte oxidant and the reductant and split the skeletal Eq (i) in to wo half reactin
Here O.N of `CI_(2)(aq) rarr2 CI^(-)(aq)`
Step 4 To balce the reduction half Equation (ii)
(a) Balance all atoms other than O and H
Since there are 2 CI atoms on L.H.S of Eq
Step 5 To balace the oxidation half Equation (iii) (a) Balance balance all atoms other thna than O and H
Since thre are 2 CI atoms on L.H.S of Eq (iii) and only one one on the R.H.S therefore multiply `CIO_(3)^(-)` ion
`CI_(2)(g) rarr 2CIO_(3)^(-)(aq)`
(b) Balance oxidation number by adding electrons the O.N of CI in `CI_(2)` o fEq (vi) is zero when on the R.H.S in `CIO_(3)^(-)` it is + 5 Thus each CI atom loss five electrons since there are two CI atoms on R.H.S therefore and `e^(-)` to R.H.S of Eq (vi) we have
`CI_(2)(g) rarr 2CIO_(3)^(-)(aq)+10 e^(-)`
Step 6 To balance the elctrons gained in eq (v) and lost in Eq (ix) multiple Eq by 5 and add to Eq (ix) we have
`5CI_(2)(g)+10 e^(-) rarr 10CI^(-)(aq)`
`CI_(2)(g)+12OH^(-)(aq) rarr 2CIO_(3)^(-)(aq)+6H_(2)O(l)+10 e^(-)`
This represent the final balanced redox equation