[anees7112]

Hello

For this question


I know how to solve it using the Nernst equation

Ecell=E0cell -0.0592/z * log10 Kc
Ecell=E0cell -0.0592/z * log10 Kc
Ecell=2.48 -0.0592/z * log10 [[0.01]*[0.01]^2/[10^-5]^4]

..where 10^-5 is the H+ concentration and is found by 10^[-ph]


My only problem is that it appears the z value is 2, but I am not sure how to work out this value.

I found this http://answers.yahoo.com/question/index?qid=20120318164407AADD11Z
but it seems like a very long method and I am not sure why he did certain steps , for example : after adding  the half reactions together he then took half reactions of the equation again.

I found the original question online http://malaysia.answers.yahoo.com/question/index?qid=20100316212226AAotBc1 and that gives n=2 .. but I am not sure how the answerer worked that out.



Thank you

[Borek]

it seems like a very long method and I am not sure why he did certain steps

It is like scratching your right ear with your left hand. Possible, but you can do it much easier with the right hand.

I found the original question online http://malaysia.answers.yahoo.com/question/index?qid=20100316212226AAotBc1 and that gives n=2 .. but I am not sure how the answerer worked that out.

Just look at the balanced reaction. Reduction is NiO₂ Ni²⁺ - two electrons are involved. Oxidation is 2Ag 2Ag⁺ - two electrons again. The second half reaction (oxidation of silver) could be simplified, but it would be no longer following stoichiometry of the oxidation that happens in the cell, so we have to keep it as 2 2.

[anees7112]

haha I liked the analogy

I get it now . Before when I tried to do that method I was also including the charge on the H+ which gave me the wrong.

Thank you for your help