# Problem: Balance the following reaction in basic solution.ClO−(aq) + Cr(OH)−4(aq) → CrO42−(aq) + Cl−(aq)A. 2OH− + 3ClO−(aq) + 2Cr(OH)4−(aq) → 2CrO42−(aq) + 3Cl−(aq) + 5H2O(l)B. 2OH− + 2ClO−(aq) + Cr(OH)4−(aq) + 2Cl−(aq) + 3H2O(l)C. 3ClO−(aq) + 2Cr(OH)4−(aq) → 2CrO42−(aq) + 3Cl−(aq) + 3H2O(l) + 2H+D. 2ClO−(aq) + Cr(OH)4−(aq) → CrO42−(aq) + Cl−(aq) + H2O(l) + 2H+

###### FREE Expert Solution

We are being asked to balance the given redox reaction occurring in a basic solution

When balancing redox reactions under basic conditions, we will follow the following steps.

Step 1: Separate the whole reaction into half-reactions
Step 2: Balance the non-hydrogen and non-oxygen elements first
Step 3: Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O)
Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)
Step 5: Balance the charges: add electrons to the more positive side (or less negative side)
Step 6: Balance electrons on the two half-reactions
Step 7: Get the overall reaction by adding the two reaction
Step 8: Balance remaining H+ by adding an equal amount of OH- ions to both sides
Step 9: H+(aq) will combine with OH-(aq) to form H2O(l)­
Step 10: Cancel out common species.

Balance the redox reaction under basic condition:

ClO(aq) + Cr(OH)4(aq) → CrO42−(aq) + Cl(aq)

84% (226 ratings) ###### Problem Details

Balance the following reaction in basic solution.

ClO(aq) + Cr(OH)4(aq) → CrO42−(aq) + Cl(aq)

A. 2OH + 3ClO(aq) + 2Cr(OH)4(aq) → 2CrO42−(aq) + 3Cl(aq) + 5H2O(l)

B. 2OH + 2ClO(aq) + Cr(OH)4(aq) + 2Cl(aq) + 3H2O(l)

C. 3ClO(aq) + 2Cr(OH)4(aq) → 2CrO42−(aq) + 3Cl(aq) + 3H2O(l) + 2H+

D. 2ClO(aq) + Cr(OH)4(aq) → CrO42−(aq) + Cl(aq) + H2O(l) + 2H+

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