Question

Balance these equations for reactions occurring in an acidic solution:

(a) NO₃^–  + Zn --> NH₄⁺ + Zn²⁺

(b) Cr³⁺ + BiO₃^– --> Cr₂O₇^2– + Bi³⁺

(c) I₂ + OCl^– --> IO₃^–  + Cl^–

(d) Mn²⁺ + BiO₃^– --> MnO₄^– + Bi³⁺

Answer 1

(a) NO₃^–  + Zn --> NH₄⁺ + Zn²⁺

first half red-ox reaction

NO3^- .........> NH4^+

add 3 H2O molecules to balance O atoms

NO3^- ..............> NH4^+ + 3H2O

to balance the H atoms add H+ ions on the left side

NO3^- + 10 H^+..............> NH4^+ + 3H2O

to balance the charge add 8e- on the left side of the above reaction

NO3^- + 10 H^+ + 8e-..............> NH4^+ + 3H2O..........................1

second half reaction

Zn ............> Zn^+2

we have to balance the charge only on the right side by adding 2e-

Zn ............> Zn^+2 + 2e- ..............2

now add eq.1 and 4*eq.2

NO3^- + 10 H^+ +4Zn..............>4Zn^+2 + NH4^+ + 3H2O

(b) Cr³⁺ + BiO₃^– --> Cr₂O₇^2– + Bi³⁺

First half red-ox reaction

Cr^+3 ...............> Cr2O7^-2

first balance the Cr atoms

2Cr^+3 ...............> Cr2O7^-2

2Cr^+3 ...............> Cr2O7^-2

add H2O molecules to balance O atoms

2Cr^+3 + 7H2O ...............> Cr2O7^-2

2Cr^+3 + 7H2O ...............> Cr2O7^-2 + 14H+

2Cr^+3 + 7H2O ...............> Cr2O7^-2 + 14H+ + 6e- ................1

second half red-ox reaction

BiO3^- .............> Bi^+3

BiO3^- .............> Bi^+3 + 3H2O

BiO3^- + 6H+ + 2e- .............> Bi^+3 + 3H2O ................2

Now add eq.1 and 3*eq.2

2Cr^+3 + 3BiO3^- + 4H+   .............> Cr2O7^-2 + 3Bi^+3 + 2H2O

(c) I₂ + OCl^– --> IO₃^–  + Cl^–

First half reaction

I2 .............> IO3^-

I2 ................> 2IO3^-

I2 + 6H2O ..............> 2IO3^-

I2 + 6H2O ..............> 2IO3^- + 12H+

I2 + 6H2O ..............> 2IO3^- + 12H+ + 10e- ...............................1

second half reaction

OCl^- ..................> Cl^-

OCl^- ....................>Cl^- + H2O

OCl^- + 2H+ + 2e-....................>Cl^- + H2O ............2

now add eq.1 and 5*eq.2

5OCl^- + H2O + I2 ....................>2IO3^- + 5Cl^- + 2H+

(d) Mn²⁺ + BiO₃^– --> MnO₄^– + Bi³⁺

First half reaction

Mn^+2 ..................> MnO4^-

Mn^+2 + 4H2O ..................> MnO4^-

Mn^+2 + 4H2O ..................> MnO4^- + 8H+

Mn^+2 + 4H2O ..................> MnO4^- + 8H+ + 5e- ....................1

second half red-ox reaction

BiO3^- .............> Bi^+3

BiO3^- .............> Bi^+3 + 3H2O

BiO3^- + 6H+ + 2e- .............> Bi^+3 + 3H2O ................2

Now add 2*eq.1 and 5*eq2

5BiO3^- +2Mn^+2 + 14H+   .............> 2MnO4^- +5Bi^+3 + 7H2O