Solving Redox Equations Involving Cl2 in Basic Solution

[jesuslovesu]

Homework Statement

Cl2 -> Cl- + ClO3-
(in basic solution)

Homework Equations

The Attempt at a Solution

Well, I've tried messing around with the half reactions but since Cl2 is involved in both, they don't seem to do much good.

e + Cl2 -> Cl-
Cl2 -> ClO3- + 5e

Normally, I would multiply the top by 5e to let the electrons cancel, but then there's a Cl2 and a 5Cl2 so that doesn't appear to work...

Anyone have any ideas on how to do this?

 

[Borek]

You are on the right track - Cl2 and a 5Cl2 is 6Cl2

Just be careful when balancing half reactions.


Equation balancer and stoichiometry calculator

 

[Blueshift5]

I would first start by balancing the equation by adding the appropriate number of electrons to each half reaction. In this case, the first half reaction would need 2 electrons and the second half reaction would need 10 electrons. This would result in the following balanced equation:

2Cl2 + 2e- -> 2Cl-
Cl2 + 5e- -> ClO3- + 6OH-

Next, I would combine the two half reactions by multiplying the first half reaction by 5 and the second half reaction by 2 to ensure that the number of electrons are equal on both sides. This would result in the final balanced equation:

5Cl2 + 5e- -> 5Cl-
2Cl2 + 10e- -> 2ClO3- + 12OH-

Finally, I would combine the two half reactions to eliminate the electrons and balance the remaining elements. In this case, the chlorine and oxygen atoms are already balanced, so the only remaining element is hydrogen. To balance hydrogen, I would add 6H+ to the left side of the equation, resulting in the final balanced equation:

5Cl2 + 5e- + 6H+ -> 5Cl- + 6H2O
2Cl2 + 10e- -> 2ClO3- + 12OH-

This balanced equation can then be used to solve for the unknown quantities, such as the concentrations of the reactants and products, using various techniques such as the Nernst equation or the Henderson-Hasselbalch equation.