Question

Complete and balance the following equations. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)

A) H₂O₂(aq) + Cl₂O₇(aq) → ClO₂⁻ (aq) + O₂(g) (basic solution)
B) Cr₂O₇^2-(aq) + I ⁻ (aq) → Cr³⁺(aq) + IO₃⁻ (aq) (acidic solution)

C) MnO₄⁻ (aq) + Cl  ⁻ (aq) → Mn²⁺(aq) + Cl₂(g) (acidic solution)

Answer 1

Step 1: Separate in half reactions.
Step 2: In each half reaction, balance all elements except O & H.
Step 3: Balance O by adding H2O.
Step 4: Balance H by adding H+.
Step 5: Balance charges by adding e-.
Step 6: Multiply all coefficients in 1 or both half reactions by an integer to get the number of e- in the two half reactions equal.
Step 7: Add the 2 half reactions together & cancel out any species that appear on both sides of net reaction.
Step 8: Check that charges and atoms are balanced.

A) So,
H2O2(aq) ---> O2(g) + 2H+ + 2e-
Cl2O7(aq) + 6H+ + 8e- ---> 2ClO2^-(aq) + 3H2O
======================================
4H2O2(aq) + Cl2O7(aq) ---> 2ClO2^-(aq) + 4O2(g) + 2H+ + 3H2O

Since it is in basic solution, add 2 OH^- to both sides, and consolidate the waters:

4H2O2(aq) + Cl2O7(aq) + 2OH^- ---> 2ClO2^-(aq) + 4O2(g) + 5H2O

B) Cr₂O₇^2- + I- + 8H⁺ -----------> 2Cr³⁺ + IO₃- + 4H₂O

C)

i.
MnO4- --> Mn2+ (reduction)
Cl- --> Cl2 (oxidation)
ii.
MnO4 --> Mn2+
2Cl- --> Cl2
iii.
MnO4- +8H+ --> Mn2+ + 4H2O
2Cl- --> Cl2
iv.
MnO4- +8H+ + 5e- --> Mn2+ + 4H2O
2Cl- --> Cl2 + 2e-
v.
2 x MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O = 2MnO4- + 16H+ + 10e- --> 2Mn2+ + 8H2O
5 x 2Cl- --> Cl2 + 2e- = 10Cl- ---> 5Cl2 + 10e-
vi.
2MnO4- + 16H+ + 10e- +10Cl- ---> 2Mn2+ + 8H2O + 5Cl2 + 10e-

Hence, Final balanced equation is
2MnO4- + 16H+ + 10Cl- ---> 2Mn2+ + 8H2O + 5Cl2