[teir95]

consider a cell constructed from two cu/cu2+ half cells, one half cell contains 0.01M CuSO4 and the other one contains 0.1M CuSO4.
write the oxidation and the reduction half reaction ?
calculate E ?

[Borek]

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You have to show your attempts at solving the question to receive help. This is a forum policy.

[teir95]

Please read the forum rules.

You have to show your attempts at solving the question to receive help. This is a forum policy.

Cu(s) ===⇒ Cu+2 (0.01 M) + 2e- (oxidation)
2e- + Cu+2(0.1 M) =⇒ Cu(s) (reduction)
---------------------------------------...
Cu(s) + Cu+2(0.1 M) =⇒ Cu+2(0.01M) + Cu(s)

This reaction occurs since the 0.10 M will become less concentrated and the 0.01 M will become more concentrated.
Nernst Equation: E = E^o – 0.059/n *logQ where Q - reaction quotient
E^o is zero and n = 2 due to a two electron change.
E = E^o – 0.059/2log(0.01)/(0.10)
E = 0.0 - 0.03log(0.10)
E = -0.03(-1)
E = +0.03

[Borek]

Looks OK to me, this is a simple concentration cell.

Note - when you are asked about the E of the cell, you don't have to trouble yourself with the sign, it is always positive, as we report the absolute value.