Question

Balance each of the following reations using the half reaction method. Show all work

a. Bi(OH)₃(S) + Sn(OH)₃^-(aq) ---> Sn(OH)₆^2- (aq) + Bi(s) (Basic solution)

b. BrO₃^-(aq) + Sb³⁺(aq) ---> Br ^- (aq) + Sb⁵⁺ (aq) (acidic solution)

Thank you for your help!

Answer 1

    +3                +2                    +4                       0

   Bi(OH)₃(S) + Sn(OH)₃^-(aq) Sn(OH)₆^2- (aq) + Bi(s)

Oxidation half reaction : Sn(OH)₃^-(aq) Sn(OH)₆^2- (aq)

Reduction half reaction : Bi(OH)₃(S) Bi(s)

Balancing oxidation half reaction :Sn(OH)₃^-(aq) Sn(OH)₆^2- (aq)

Balancing O atoms ( by adding H₂O ) : Sn(OH)₃^-(aq) + 3H₂O Sn(OH)₆^2- (aq)

Balancing H atoms ( by adding H₂O & OH^- ) : Sn(OH)₃^-(aq) + 3H₂O + 3OH^- Sn(OH)₆^2- (aq)+ 3H₂O

Balancing charge ( by adding electrons) : Sn(OH)₃^-(aq) + 3H₂O + 3OH^- Sn(OH)₆^2- (aq)+ 3H₂O + 2e^-

  Sn(OH)₃^-(aq) + 3OH^- Sn(OH)₆^2- (aq) + 2e^-                ---(1)

Balancing Reduction half reaction : Bi(OH)₃(S) Bi(s)

Balancing O atoms ( by adding H₂O ) : Bi(OH)₃(S) Bi(s) + 3H₂O

Balancing H atoms ( by adding H₂O & OH^- ) : Bi(OH)₃(S) +3H₂O Bi(s) + 3H₂O + 3OH^-

Balancing charge ( by adding electrons) : Bi(OH)₃(S) +3H₂O+ 3e^- Bi(s) + 3H₂O + 3OH^-     

               Bi(OH)₃(S) + 3e^- Bi(s) + 3OH^-      ---(2)

Adding both the equations so that the number of electrons must be balanced on both sides.It can be done as follows:

[3xEqn(1) ] + [2xEqn (2)]   gives

3Sn(OH)₃^-(aq) + 9OH^-+ 2Bi(OH)₃(S) + 6e^-   3Sn(OH)₆^2- (aq)+ 6e^- +2Bi(s) + 6OH^-  

3Sn(OH)₃^-(aq) + 3OH^-+ 2Bi(OH)₃(S)   3Sn(OH)₆^2- (aq)+ 2Bi(s)

This is the balance equation.

-----------------------------------------------------------------------------------------------------

   +5               +3               -1           +5

   BrO₃^-(aq) + Sb³⁺(aq) Br ^- (aq) + Sb⁵⁺ (aq)

Oxidation half reaction : Sb³⁺(aq) Sb⁵⁺ (aq)

Reduction half reaction : BrO₃^-(aq) Br ^- (aq)

Balancing oxidation half reaction :Sb³⁺(aq) Sb⁵⁺ (aq)

Balancing charge ( by adding electrons) : Sb³⁺(aq) Sb⁵⁺ (aq) + 2e^-           ---(1)

Balancing Reduction half reaction : BrO₃^-(aq) Br ^- (aq)

Balancing O atoms ( by adding H₂O ) : BrO₃^-(aq) Br ^- (aq) + 3H₂O

Balancing H atoms ( by adding H⁺ ) : BrO₃^-(aq) + 6H⁺ Br ^- (aq) + 3H₂O

Balancing charge ( by adding electrons) : BrO₃^-(aq) + 6H⁺ +6e^- Br ^- (aq) + 3H₂O                   ---(2)

Adding both the equations so that the number of electrons must be balanced on both sides.It can be done as follows:

[3xEqn(1) ] + [1xEqn (2)]   gives

3Sb³⁺(aq)+ BrO₃^-(aq) + 6H⁺ +6e^- 3Sb⁵⁺ (aq) + 6e^-+ Br ^- (aq) + 3H₂O        

3Sb³⁺(aq)+ BrO₃^-(aq) + 6H⁺ 3Sb⁵⁺ (aq) + Br ^- (aq) + 3H₂O

This is the balance equation.