Question

Balance the following redox equations either in acid or base:

a. HClO -> Cl^(-) + ClO3^(-) [acid]

b. ClO4^(-) + Br2 -> Cl^(-) + BrO2^(-) [base]

c. H2S + BrO3^(-) -> SO3^(2-) + Br2 [base]

d. NH3 + SO4^(2-) -> N2 + SO^(2) [acid]

e. IO3^(-) + H2S -> I2 + SO4^(2-) [acid]

Answer 1

(a) HClO ---> Cl^- + ClO₃^- in acidic medium :

First Step : Separating the two half cell reactions :

HClO ---> Cl-

HClO ----> ClO3-

Next step : Balancing elements other than O and H :

HClO ---> Cl-

HClO ----> ClO3-

Next step : Adding H2O to balance O :

HClO ---> Cl- + H2O

HClO + 2H2O ----> ClO3-

Next step : Balancing H with protons :

HClO + H+ ---> Cl- + H2O

HClO + 2H2O ----> ClO3- + 5H+

Next step : Balancing charge with e- :

HClO + H+ + 2e- ---> Cl- + H2O

HClO + 2H2O ----> ClO3- + 5H+ + 4e-

Next step : Scaling reactions so that they have equal amount of e- :

2HClO + 2H+ + 4e- ---> 2Cl- + 2H2O

HClO + 2H2O ----> ClO3- + 5H+ + 4e-

Final step : Adding reactions and cancelling e- and other common terms:

2HClO + HClO ---> 2Cl^- + ClO₃^- + 3H⁺

(b) ClO4^(-) + Br2 -> Cl^(-) + BrO2^(-) [base] :

First Step : Separating the two half cell reactions :

ClO4- ---> Cl-

Br2 ----> BrO2-

Next step : Balancing elements other than O and H :

ClO4- ---> Cl-

Br2 ----> 2BrO2-

Next step : Adding H2O to balance O :

ClO4- ---> Cl- + 4H2O

Br2 + 2H2O ----> 2BrO2-

Next step : Balancing H with protons :

ClO4- + 8H+---> Cl- + 4H2O

Br2 + 2H2O ----> 2BrO2- + 4H+

Next step : Balancing charge with e- :

ClO4- + 8H+ + 8e- ---> Cl- + 4H2O

Br2 + 2H2O ----> 2BrO2- + 4H+ + 2e-

Next step : Scaling reactions so that they have equal amount of e- :

ClO4- + 8H+ + 8e- ---> Cl- + 4H2O

4Br2 + 8H2O ----> 8BrO2- + 16H+ + 8e-

Next step : Adding reactions and cancelling e- and other common terms:

ClO4- + 4Br2 + 8H2O ---> Cl- + 4H2O + 8BrO2- + 8H+

Next step : Add OH- to balance protons :

ClO4- + 4Br2 + 8H2O + 8OH- ---> Cl- + 4H2O + 8BrO2- + 8H+ + 8OH-

Final step : Combining OH- and H+ to form water and cancelling common terms :

ClO₄^- + 4Br₂ + 8OH^- ---> Cl^- + 8BrO₂^- + 4H₂O

(c) Required balanced reaction in basic medium is :

5H₂S + 4 OH^- + 6BrO₃^- ---> 5 SO₃^2- + 3Br₂ + 7H₂O

(d) NH3 + SO4^(2-) -> N2 + SO^(2) [acid] :

2NH₃ + 6H⁺ + 3SO₄^2- ----> N₂ + 3SO₂ + 6H₂O

(e) IO3^(-) + H2S ---> I2 + SO4^(2-) [acid] :

8IO₃^- + 5H₂S -0---> 4I2 + 4H₂O + 5SO₄^2- + 2H⁺