Question

In: Chemistry

Balance each of the following reations using the half reaction method. Show all work a. Bi(OH)3(S)...

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!

Expert Solution

+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.

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