In: Chemistry
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)62- (aq) + Bi(s) (Basic solution)
b. BrO3-(aq) + Sb3+(aq) ---> Br - (aq) + Sb5+ (aq) (acidic solution)
Thank you for your help!
+3 +2 +4 0
Bi(OH)3(S) + Sn(OH)3-(aq) Sn(OH)62- (aq) + Bi(s)
Oxidation half reaction : Sn(OH)3-(aq) Sn(OH)62- (aq)
Reduction half reaction : Bi(OH)3(S) Bi(s)
Balancing oxidation half reaction :Sn(OH)3-(aq) Sn(OH)62- (aq)
Balancing O atoms ( by adding H2O ) : Sn(OH)3-(aq) + 3H2O Sn(OH)62- (aq)
Balancing H atoms ( by adding H2O & OH- ) : Sn(OH)3-(aq) + 3H2O + 3OH- Sn(OH)62- (aq)+ 3H2O
Balancing charge ( by adding electrons) : Sn(OH)3-(aq) + 3H2O + 3OH- Sn(OH)62- (aq)+ 3H2O + 2e-
Sn(OH)3-(aq) + 3OH- Sn(OH)62- (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)62- (aq)+ 6e- +2Bi(s) + 6OH-
3Sn(OH)3-(aq) + 3OH-+ 2Bi(OH)3(S) 3Sn(OH)62- (aq)+ 2Bi(s)
This is the balance equation.
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+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.