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Question
Balance the following reaction by oxidation number method.
\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)} + Sn(OH)^2-_{6(aq)}(basic)}\]
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Solution
\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)} + Sn(OH)^2-_{6(aq)}(basic)}\]
Step 1: Write the skeletal equation and balance the elements other than O and H.
\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)} + Sn(OH)^2-_{6(aq)}}\]
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.
Increase in oxidation number:
(Increase per atom = 2)
Decrease in oxidation number:
(Decrease per atom = 3)
To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)}}\]
Step 3: Balance ‘O’ atoms by adding 3H2O to the left-hand side.
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3H2O_{(l)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)}}\]
Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3H+ on the right-hand side.
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3H2O_{(l)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)} + 3H^+_{( aq)}}\]
Add OH− ions equal to the number of H+ ions on both sides of the equation.
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3H2O_{(l)} + 3OH^-_{( aq)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)} + 3H^+_{( aq)} + 3OH^-_{( aq)}}\]
The H+ and OH− ions appearing on the same side of the reaction are combined to give H2O molecules.
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3H2O_{(l)} + 3OH^-_{( aq)}->4Bi_{(s)} + 3Sn(OH)^2-_{6(aq)} + 3H2O_{(l)}}\]
\[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3OH^-_{( aq)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)}}\]
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \[\ce{2Bi(OH)_{3(s)} + 3Sn(OH)^-_{3(aq)} + 3OH^-_{( aq)}->2Bi_{(s)} + 3Sn(OH)^2-_{6(aq)}}\]
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