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From the equation deltaG = -nFE, you can tell that a reaction is spontaneous and does work if the value of Ecell > 0. Since to find E you would subtract the potential of the cathode by the potential of the anode, you will always get Ecell > 0 if the cathode value is larger.
Because if there is still potential difference between the cathode and anode, then the reaction is not at equilibrium, and electrons will continue to flow from the cathode to the anode. This generates electricity.
This is because the Evalue at the cathode basically determines if the Ecell will be positive or negative. When the value is positive the system can do work, when the value is negative work must be done on the system. The equation is Ecell=Ecath-Eannode, so if Ecath>Eanode, the equation will have a net positive value.
On the right, a cathode is part of the reduction half reaction. On the left, an anode is part of the oxidation reaction. Since overall potential and E is the difference of the potential of the right, minus the potential of the left, useful work can be done when the cathode os greater than the anode.
When E(cell) is not = 0, the cell can do work. Since E(cell) = E(cathode) - E(anode), if E(cell) is > 0, that means E(cathode) > E(anode) and the cell will do work. I assume that E(cell) can also be < 0 and do work, and in this case E(anode) > E(cathode)?