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

Balance the following by ion electron method (basic medium): `Cr(OH)_(

1.	Balance the following by ion electron method (basic medium): `Cr(OH)_(3)+IO_(3)^( ө)rarrI^(ө)+CrO_(4)^(2-)`
Answer» Two half reactions are: a. `Cr(OH)_(3)rarrCeO_(4)^(2-)` b. `IO_(3)^(ө)rarrI^(ө)` Balancing `O` atoms: a. `Cr(OH)_(3)+2overset(ө)OHrarrCrO_(4)^(2-)+H_(2)O` b. `IO_(3)^(ө)+3H_(2)OrarrI^(ө)+6overset(ө)OH` Balancing `H` atoms: a. `Cr(OH)_(3)+2overset(ө)OH+3overset(ө)OHrarrCrO_(4)^(2-)+H_(2)O+3H_(2)O` b. `IO_(3)^(ө)+3H_(2)OrarrI^(ө)+6overset(ө)OH` Balancing the charge: a. `Cr(OH)_(3)5overset(ө)OHrarrCrO_(4)^(2-)+4H_(2)O^(+)3e^(-)` b. `IO_(3)^(ө)+3H_(2)O+6e^(-)rarrI^(ө)+6overset(ө)OH` Adding (a) and (b), we get `2Cr(OH)_(3)+IO_(3)^(ө)+4overset(ө)OHrarr2CrO_(4)^(2-)+I^(ө)+5H_(2)O`

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

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