I don't think that's right. I think it will be easier if you try to think physically and apply Le Chatelier's principle.

Standard electrode potentials are defined for the reduction half-reaction. The electrode potential will be increased by any change that tends to push the reaction to the right, e.g. an increase in concentration of the oxidised form or a reduction in concentration of the reduced form.

So E = E° + RT/nF*ln([ox]/[rd]) {Insert exponents as required by stoichiometry, but you get the point I hope}

This applies to both cathode and anode. Then

E

_{cell} = E

_{c} - E

_{a} = E

_{c}° - E

_{a}° + RT/nF*ln([ox]

_{c}[rd]

_{a}/[rd]

_{c}[ox]

_{a})

= E

_{cell}° - RT/nF*lnQ

where Q is the reaction quotient for the overall reaction ox

_{c} + rd

_{a} rd

_{c} + ox

_{a}The negative sign is because Q has products on top and reactants on the bottom, so an increase in Q would tend to push the reaction to the left, and so reduce the cell emf.

You are getting confused with your anode potentials. Your statement that E

_{a}=E

_{a}^{0}-R*T/z*F*ln([M

_{a}^{z+}]/[M

_{a}]) is correct if E

_{a}° is the potential for the

*oxidation*, i.e.

* minus * the standard reduction potential. But then E

_{cell} = E

_{c} + E

_{a} (as so defined). In short, you can regard the overall reaction as the cathodic reduction

**plus** the anodic oxidation, or as the cathodic reduction

**minus** the anodic reduction, but

**not** as the reduction minus the oxidation.

This all looks complicated, but I hope it helps to clarify things a bit. Briefly, I suggest:

(i) Think in terms of reduction half-reactions and potentials, subtract as appropriate (anode from cathode).

(ii) Apply Le Chatelier.