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In passing chlorine gas through a concentrated solution of alkali we get chloride and chlorate ions obtain balaced chemical equatoin for this reaction |
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Answer» Solution :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 CIin `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 |
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