In the reaction
`xCl_2 + yOH^(-) rarr Cl^(-) + ClO_3^(-)` .

Options

A. `x=2, y=4`
B. ` x=3,y=6`
C. ` x=4, y=8`
D. ` x=5, y=10`

Answer

B

Explanation

.
Total increase of oxidation number ` =2 xx 5 = 10`
Total decrease of oxidation number ` = 2xx -1 =- 2`
Here, `Cl_2` acts as both an oxidizing agent as well as reducing agent . Thus ,
Oxidation half-reaction
` Cl_2 rarr ClO_3^(-)`
Reduction half-reaction
` Cl_2 rarr Cl^(-)`
Now , we balance the two half-reactons separately
(a) Balance all atoms except H and O :
` Cl_2 rarr 2ClO_3^(-)`
` Cl_2 rarr 2 Cl^(-)`
(b) Account for the change in oxidation number
`overset (0) (C)l_2 rarr 2overset (+5) (Cl)O_(3)^(-) + 10e^(-)`
`overset (0) (C)l_2 + 2e^(-) rarr 2overset (-2)(Cl^(-))`
(c ) Balance ionic charges by adding ` OH^(-)` ions as the reaction is carried out in the basic medium :
` Cl_2 + 12OH^(-) rarr 2 ClO_3^(-) + 10e^(-)`
` Cl_2+ 2e^(-) rarr 2Cl^(-)`
(d) Balance H and O by adding ` H_2 O` molecules
` Cl_2 + 12OH^(-) rarr 2ClO_3^(-)n + 6H_2O + 10e^(-)`
` Cl_2 + 2e^(-)rarr 2Cl^(-) `
To equalize the number of electrons lost and gaind multiply reduction half-reaction by 5. Finally add them, cancelling electrons on both sides:
`6Cl_2 + 12OH^(-) rarr 10 Cl^(-) = 2ClO_3^(-)+ 6H_2 O`
or `3Cl_2 + 6O H^(-) rarr 5Cl^(-)+ClO_(3)^(-) + 3H_2 O`