Balancing Redox Reaction in Basic Solution: Step-by-Step Guide with Examples

Balance the redox reaction below if it takes place in a basic solution.Cr(OH)4- (aq) + ClO- (aq) → CrO42- (aq) + Cl- (aq)

2OH^-(aq)+2Cr(OH)4^-(aq)+3ClO^-(aq)=2CrO4^2-(aq)+5H2O(l)+3Cl^-(aq)

Before balancing the given redox reaction, we need to recognize that it is taking place in a basic solution. This means we need to add OH- ions to both sides of the reaction to balance the hydrogen and oxygen atoms.

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

Next, we will separate the reaction into half-reactions, one for oxidation and the other for reduction:

Oxidation half-reaction: Cr(OH)4- (aq) → CrO42- (aq)

Reduction half-reaction: ClO- (aq) → Cl- (aq)

Now, we will balance each half-reaction separately:

Oxidation half-reaction:

Cr(OH)4- (aq) → CrO42- (aq)
Adding water to balance the oxygen atoms:
Cr(OH)4- (aq) → CrO42- (aq) + 2H2O
Adding electrons to balance the charge:
Cr(OH)4- (aq) → CrO42- (aq) + 2H2O + 8e-

Reduction half-reaction:

ClO- (aq) → Cl- (aq)
Adding water to balance the oxygen atoms:
ClO- (aq) + H2O → Cl- (aq) + 2OH-
Adding electrons to balance the charge:
ClO- (aq) + H2O + 2e- → Cl- (aq) + 2OH-

Now we need to balance the number of electrons in both half-reactions. To do this, we need to multiply the oxidation half-reaction by 2:

2Cr(OH)4- (aq) → 2CrO42- (aq) + 4H2O + 16e-

Now both half-reactions have 16 electrons.

To get the final balanced equation, we add the two half-reactions together and cancel out common terms to get the net equation:

2Cr(OH)4- (aq) + ClO- (aq) + H2O → 2CrO42- (aq) + Cl- (aq) + 4OH-

This is the balanced redox reaction in a basic solution.

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