Balance Half Reactions: Oxidation and Reduction

Question

1. Cu+2(aq) + K (s) ---> Cu (s) + K+(aq) 

It tells me to write the balanced half reactions for oxidation and reduction. Then balance the overall equation.

Please help, it would be greatly appreciated because I am stuck and have no clue what to do. Thanks.

Answer 1

Cu⁺²(aq) + K (s) ---> Cu (s) + K+(aq) 

** Cu on the left (Cu²⁺) goes to Cu on the right. So it has GAINED 2 electrons, meaning it was reduced.

** K on the left goes to K+ on the right. It has lost 1 electrons meaning it was oxidized.

Half reactions:

Cu²⁺ + 2e- ==> Cu(s) ... balanced reduction half reaction

K(s) ===> K⁺ + 1e- ... balanced oxidation half reaction

If you want the overall balanced reaction (which wasn't asked for in the problem), you would have to multiply the oxidation half reaction by 2 in order to equalize the number of electrons transferred. Then you would add the two reactions together to obtain...

Cu²⁺ + 2K(s) ==> Cu(s) + 2K⁺ ... balanced overall reaction

Answer 2

Hi Alex J.,

A half-reaction is just a "theoretical piece" of your total reaction here. It shows just one of the elemental species (which can be the metal, or nonmetal, or oxyanions of them, etc.) being reduced or oxidized (you can show it either direction, tables of these always show them in a standard direction, oxidation I think). So, for example,

Cu⁺² + 2 e^- = 2 Cu⁰ which you could also write just as well, as:

(1/2) Cu⁺² + e^- = Cu⁰ ----- because the oxidation or reduction reaction doesn't care how much stuff is reacting, the electrode potential is an intensive quantity!

These are shown in the reduction direction, just switch the two sides around for the oxidation direction.

In some cases, half-reactions also include water being split to provide/consume H+, or O for/from complex species formation/destruction. Just sayin'.

I'll leave it to you to discover the other, similar "piece" for K . Then, to balance the total reaction, you just need to scale up each "half-reaction" as needed until the two pieces match in terms of electrons produced and consumed respectively. Here, it's easy to see that twice as much K (on a molar basis) is needed as Cu, since each K is only half as effective about producing electrons (one for K, two for Cu->Cu⁺² ).

-- Cheers, --Mr. d.