To balance the redox reaction in a basic solution, follow these detailed steps:
1. Separate the reaction into two half-reactions (oxidation and reduction):
[tex]\[
\begin{align*}
\text{Oxidation half-reaction:} & \quad Cr(OH)_3 \rightarrow CrO_4^{2-}\\
\text{Reduction half-reaction:} & \quad Cl_2 \rightarrow Cl^-
\end{align*}
\][/tex]
2. Balance each half-reaction for atoms other than hydrogen (H) and oxygen (O):
For the oxidation half-reaction:
- Chromium (Cr) atoms are already balanced.
For the reduction half-reaction:
[tex]\[
Cl_2 \rightarrow 2Cl^-
\][/tex]
Chlorine (Cl) atoms are balanced.
3. Balance oxygen atoms by adding water (H₂O):
For the oxidation half-reaction:
[tex]\[
Cr(OH)_3 \rightarrow CrO_4^{2-} + 4H_2O
\][/tex]
4. Balance hydrogen atoms by adding hydroxide ions (OH⁻) since it is a basic solution:
For the oxidation half-reaction:
[tex]\[
Cr(OH)_3 \rightarrow CrO_4^{2-} + 4H_2O
\][/tex]
Since we need to balance the H atoms, we add the required OH⁻ and water. This results in adding 6OH⁻ to the product side to neutralize the H atoms from water. This step also introduces the formula for water in the product side:
[tex]\[
Cr(OH)_3 + 6OH^- \rightarrow CrO_4^{2-} + 3H_2O
\][/tex]
5. Balance the charges by adding electrons:
For the oxidation half-reaction:
[tex]\[
Cr(OH)_3 + 6OH^- \rightarrow CrO_4^{2-} + 3H_2O + 3e^-
\][/tex]
The reduction half-reaction:
[tex]\[
Cl_2 + 2e^- \rightarrow 2Cl^-
\][/tex]
6. Make the electron loss equal to the electron gain:
Multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance electrons:
Oxidation half-reaction:
[tex]\[
2(Cr(OH)_3 + 6OH^- \rightarrow CrO_4^{2-} + 3H_2O + 3e^-)
\][/tex]
This gives:
[tex]\[
2Cr(OH)_3 + 12OH^- \rightarrow 2CrO_4^{2-} + 6H_2O + 6e^-
\][/tex]
Reduction half-reaction:
[tex]\[
3Cl_2 + 6e^- \rightarrow 6Cl^-
\][/tex]
7. Add the two balanced half-reactions together:
Combining both half-reactions:
[tex]\[
2Cr(OH)_3 + 3Cl_2 + 12OH^- \rightarrow 2CrO_4^{2-} + 6Cl^- + 6H_2O
\][/tex]
8. Simplify the equation if needed:
In this step, we simplify the equation where possible.
The final balanced equation in a basic medium is:
[tex]\[
2Cr(OH)_3 (s) + 3Cl_2 (g) + 6OH^- (aq) \rightarrow 2CrO_4^{2-} (aq) + 6Cl^- (aq) + 6H_2O (l)
\][/tex]
