What is the easiest way to calculate E cell values?

This is a calculation which students often find tricky. One fail-safe method for doing this calculation is considering E cell = E (reduced) - E(Oxidised). A more positive electrode potential means that a species is more easily able to gain electrons (and therefore more easily reduced). Conversely, a more negative electrode potential means that a species is more easily able to give away electrons (and so more easily oxidised).
Consider the example below:
Zn²⁺_(aq) + 2e^- --> Zn _(s) E^O = - 0.76 V
Cu²⁺_(aq) + 2e^- --> Cu_(s) E^O = +0.34 VIf we apply the principles outlined above, it is the Copper species which will be reduced - as it has the more positive reduction potential. We need to work out which Copper species is reduced i.e. is it Cu²⁺_(aq) or Cu_(s)?In this case, it must be the former as the copper half equation shows that only Cu2+ can gain electrons, but not Cu (s).
This must mean that the Zinc Species is oxidised - as it has the more negative reduction potential. In this case, Zn (s) must be oxidised as it is the only zinc species which can lose electrons according to this half-equation.
Therefore, E cell = E reduced - E oxidised = + 0.34 V - (-0.76 V) = + 1.10 V. As the E cell value calculated is positive, the reaction is thermodynamicaly spontaneous!
Therefore, the equation for the reaction is: Zn_(s) + Cu²⁺_(aq) --> Zn²⁺ _(aq) + Cu _(s).Bear in mind, the reaction might not occur if it has a high activation energy, meaning that there will be a low rate of reaction i.e. it is kinetically unfavourable.