Answer to Question #87380 in Chemistry for Florence

Question

Balance the following reactions
a. Cu(NH3)4^2 + S2O4^2- ==> SO3^2- + Cu + NH3(in acidic solution)
b. I- + MnO4- ==> I2 + MnO2 (in basis solution)

Answer 1

Task:

Balance the following reactions

a. Cu(NH₃)₄²⁺ + S₂O₄^2- ==> SO₃^2- + Cu + NH₃ (in acidic solution)

b. I^- + MnO₄^- ==> I₂ + MnO₂ (in basic solution)

Solution (a):

Cu(NH₃)₄²⁺ + S₂O₄^2- ==> SO₃^2- + Cu + NH₃ (in acidic solution)???

Cu(NH₃)₄²⁺ + S₂O₄^2- ==> SO₃^2- + Cu + NH₃ (in basic solution)

In basic medium:

(S₂O₄^2- + 4OH^- - 2e(-) ----> 2SO₃^2- + 2H₂O) * 1 - oxidation

(Cu(NH₃)₄²⁺ + 2e(-) ----> Cu + 4NH₃) *1 – reduction

S₂O₄^2- + 4OH^- - 2e(-) + Cu(NH₃)₄²⁺ + 2e(-) ----> 2SO₃^2- + 2H₂O + Cu + 4NH₃

S₂O₄^2- + 4OH^- + Cu(NH₃)₄²⁺ ----> 2SO₃^2- + 2H₂O + Cu + 4NH₃

Answer (a): S₂O₄^2- + 4OH^- + Cu(NH₃)₄²⁺ ----> 2SO₃^2- + 2H₂O + Cu + 4NH₃

Solution (b):

I^- + MnO₄^- ==> I₂ + MnO₂ (in basic solution)???

In basic medium:

MnO₄^-+ e(-) ----> MnO₄^2-

I^- + MnO₄^- ==> I₂ + MnO₂ (in neutral solution)

In neutral medium:

(2H₂O + MnO₄^- + 3e(-) ----> MnO₂ + 4OH^-) * 2 - reduction

(2I^- - 2e(-) ----> I₂ ) * 3 - oxidation

4H₂O + 2MnO₄^- + 6e(-) + 6I^- - 6e(-) ----> 2MnO₂ + 8OH^- + 3I₂

4H₂O + 2MnO₄^- + 6I^-  ----> 2MnO₂ + 8OH^- + 3I₂

Answer (a): 4H₂O + 2MnO₄^- + 6I^-  ----> 2MnO₂ + 8OH^- + 3I₂