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Balance the following by ion electron me...

Balance the following by ion electron method (basic medium):
`Cr(OH)_(3)+IO_(3)^( ө)rarrI^(ө)+CrO_(4)^(2-)`

Text Solution

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`2Cr(OH)_(3) + IO_(3)^(-) to I^(-) + 2CrO_(4)^(2-)`
Balancing of O and H
`2Cr(OH)_(3) + IO_(3)^(-) + 4OH^(-) to I^(-) + 2CrO_(4)^(2-) + H_(2)O + 4H_(2)O`
Now,
`2Cr(OH)_(3)+IO_(3)^(-) + 4OH^(-) to I^(-) + 2CrO_(4)^(2-) + 5H_(2)O`
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