Question

Balance the following aqueous reactions:

a. CrI₃ + Cl₂ -----> CrO₄^2- + IO₄^1- + Cl^1- (basic solution)

b. Cr(NCS)₆^4- + Ce⁴⁺ ----> Cr³⁺ + Ce³⁺ + NO₃^1- + CO₂ + SO₄^2- (acidic solution)

Answer 1

Increase in oxidation number (O.N) is known as oxidation. Decrease in oxidation number is known as reduction.

a. Oxidation half reaction

CrI3 + 16 H2O -------> CrO4^2- + 3 IO4^- + 32 H^+ + 27e^- ------(1)

Reduction half reaction

Cl2 + 2e^- -----> 2Cl^- --------(2)

Multiply (1) by 2 and (2) by 27 and then add it

2CrI3 + 32 H2O ----> 2CrO4^2- + 6 IO4^- + 64 H^+ + 54 e^-

27 Cl2 + 54 e^- -----> 54 Cl^-  

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

2CrI3 + 27 Cl2 + 32 H2O ---->2CrO4^2- + 6IO4^- + 54Cl^- + 64 H^+

Add 64 OH^- both sides to neutralize 64 H^+

2CrI3 + 27 Cl2 + 32 H2O + 64 OH^- ---> 2CrO4^2- + 6IO4^- + 54 Cl^- + 64 H^+ + 64 OH^-

2CrI3 + 27 Cl2 + 32 H2O + 64 OH^- ----> 2CrO4^2- + 6IO4^- + 54 Cl^- + 64 H2O

2CrI3 + 27 Cl2 + 64 OH^- -----> 2CrO4^2- + 6IO4^- + 54 Cl^- + 32 H2O

b. Oxidation half reaction

Cr(NCS)6^4- + 54 H2O ----> Cr^3+ +6 NO3^- + 6CO2 + 6SO4^2- + 108 H^+ 97 e^- -----(3)

Reduction half reaction

Ce^4+ + e^- -----> Ce3+ -----(4)

Multiply (4) by 97 and add to (3)

Cr(NCS)6^4- + 54 H2O ---> Cr^3+ + 6NO3^- + 6CO2 + 6SO4^2- + 108 H^+ + 97 e^-

97 Ce^4+ + 97 e^- ----> 97 Ce^3+

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

Cr(NCS)6^4- + 97 Ce^4+ + 54 H2O -----> 97 Ce^3+ + Cr^3+ + 6NO3^- + 6CO2 + 6SO4^2- + 108 H^+