Question

The value of the equilibrium constant for the electrochemical reaction

MnO₄^–_(aq) + 2H⁺_(aq) + ½Cl_2(g) ⇌Mn²⁺_(aq)+ ClO₃^–_(aq) +H₂O_(aq)

is 5.88 x10^–5. What is the value of the equilibrium constant for the reverse reaction?

Answer 1

1 answer · Chemistry

Best Answer

First, split up the reaction into half-reactions. To do this, simply use the elements that are oxidised or reduced, but not the other. In this case, those elements are Mn and Cl, so make a separate equation for each. Ignore states for now, you can add them in later.

MnO4^- → Mn^2+
Cl^- → Cl2

Now you have to balance them. Balance the elements and electrons, balance O by adding H2O, and H by adding H^+. You can balance the electrons at the end by checking the total charge on each side, then adding electrons to the more positive side until the overall charges balance.

MnO4^- → Mn^2+
MnO4^- → Mn^2+ + 4H2O (balance O)
MnO4^- + 8H^+ → Mn^2+ + 4H2O (balance H)

The left side has a total charge of 8-1 = +7, and the right has a total of +2. To balance this, add 5 electrons to the left side (i.e. 5 negative charges).

MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O

This part is now balanced. Now do the Cl equation, which is a lot easier.

Cl^- → Cl2
2Cl^- → Cl2
2Cl^- → Cl2 + 2e^-

Now you have to combine the equations so that the electrons cancel. 5 electrons in the Mn equation and 2 electrons in the Cl equation can be balanced by taking twice the Mn equation (for a total of 10 electrons) and 5 times the Cl equation (for a total of 10). Then add the rest, and add the states again.

2MnO4^- + 16H^+ + 10e^- → 2Mn^2+ + 8H2O
10Cl^- → 5Cl2 + 10e^-
———————————————————————————————
2MnO4^- + 16H^+ + 10e^- + 10Cl^- → 2Mn^2+ + 8H2O + 5Cl2 + 10e^-
2MnO4^-(aq) + 16H^+(aq) + 10Cl^-(aq) → 2Mn^2+(aq) + 5Cl2(aq) + 8H2O(l)

This is now balanced.