Question

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which side of the reaction do they appear?

MnO41- (aq) + Cl1- (aq) → Mn2+ (aq) + Cl2 (g)

a. 2 moles of H2O on the reactant side
b. 2 moles of H2O on the product side
c. 4 moles of H2O on the product side
d. 8 moles of H2O on the product side
e. 10 moles of H2O on the reactant side

Answer 1

d. 8 moles of H2O on the product side

Step-by-step explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

`MnO_4^(1-) (aq) + Cl^(1-) (aq) \rightarrow Mn^(2+) (aq) + Cl_2 (g)\\\\(Mn^(7+)O^(2-)_4)^(1-) (aq) + Cl^(1-) (aq) \rightarrow Mn^(2+) (aq) + Cl_2 (g)\\\\\\\\(Mn^(7+)O^(2-)_4)^(1-) (aq)+8H^++5e^- \rightarrow Mn^(2+)+4H_2O\\\\2Cl^(1-)\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^(7+)O^(2-)_4)^(1-) (aq)+8H^++5e^- \rightarrow Mn^(2+)+4H_2O]\\\\5*[2Cl^(1-)\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^(7+)O^(2-)_4)^(1-) (aq)+16H^++10e^- \rightarrow 2Mn^(2+)+8H_2O\\\\10Cl^(1-)\rightarrow 5Cl_2^0+10e^-\\`

Then, we add the half reactions:

`2(Mn^(7+)O^(2-)_4)^(1-) (aq)+16H^++10Cl^(1-) \rightarrow 2Mn^(2+)+8H_2O+5Cl_2^0`

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.