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Question
Balance the following redox equation by half-reaction method.
\[\ce{Bi(OH)_{3(s)} + SnO^2-_{2(aq)}->SnO^2-_{3(aq)} + Bi^_{(s)}(basic)}\]
Solution
\[\ce{Bi(OH)_{3(s)} + SnO^2-_{2(aq)}->SnO^2-_{3(aq)} + Bi^_{(s)}}\]
Step 1: Write the unbalanced equation for the redox reaction. Assign the oxidation number to all the atoms in reactants and products.
Divide the equation into two half equations.
- Oxidation half-reaction: \[\ce{SnO^2-_{2(aq)}->SnO^2-_{3(aq)}}\]
- Reduction half-reaction: \[\ce{Bi(OH)_{3(s)}->Bi_{(s)}}\]
Step 2: Balance half equations for \[\ce{O}\] atoms by adding \[\ce{H2O}\] to the side with fewer \[\ce{O}\] atoms. Add 1 \[\ce{H2O}\] to the left side of the oxidation half equation and \[\ce{3H2O}\] to the right side of the reduction half equation.
- Oxidation: \[\ce{SnO^2-_{2(aq)} + H2O_{(l)}->SnO^2-_{3(aq)}}\]
- Reduction: \[\ce{Bi(OH)_{3(s)}->Bi_{(s)} + 3H2O_{(l)}}\]
Step 3: Balance \[\ce{H+}\] atoms by adding \[\ce{H+}\] ions to the side with less \[\ce{H}\]. Hence, add \[\ce{2H+}\] ions to the right side of the oxidation half equation and \[\ce{3H+}\] ions to the left side of the reduction half equation.
- Oxidation: \[\ce{SnO^2-_{2(aq)} + H2O_{(l)}->SnO^2-_{3(aq)} + 2H^+_{( aq)}}\]
- Reduction: \[\ce{Bi(OH)_{3(s)} + 3H^+_{( aq)}->Bi_{(s)} + 3H2O_{(l)}}\]
Step 4: Now add 2 electrons to the right side of the oxidation half equation and 3 electrons to the left side of the reduction half equation to balance the charges.
- Oxidation: \[\ce{SnO^2-_{2(aq)} + H2O_{(l)}->SnO^2-_{3(aq)} + 2H^+_{( aq)} + 2e-}\]
- Reduction: \[\ce{Bi(OH)_{3(s)} + 3H^+_{( aq)} + 3e- ->Bi_{(s)} + 3H2O_{(l)}}\]
Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize the number of electrons in two half equations.
Then add two half equations.
- Oxidation: \[\ce{3SnO^2-_{2(aq)} + 3H2O_{(l)}->3SnO^2-_{3(aq)} + 6H^+_{( aq)} + 6e-}\]
- Reduction: \[\ce{2Bi(OH)_{3(s)} + 6H^+_{( aq)} + 6e- ->2Bi_{(s)} + 6H2O_{(l)}}\]
Add two half equations:
\[\ce{2Bi(OH)_{3(s)} + 3SnO^2-_{2(aq)}->3SnO^2-_{3(aq)} + 2Bi_{(s)} + 3H2O_{(l)}}\]
Reaction occurs in basic medium. However, \[\ce{H+}\] ions cancel out and the reaction is balanced. Hence, no need to add \[\ce{OH-}\] ions. The equation is balanced in terms of number of atoms and the charges
Hence, balanced equation: \[\ce{2Bi(OH)_{3(s)} + 3SnO^2-_{2(aq)}->3SnO^2-_{3(aq)} + 2Bi_{(s)} + 3H2O_{(l)}}\]
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