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For emample, if you are dealing with Copper and Zinc, you first take the half equations Cu2+ + 2e- --> Cu +0.34 Zn2+ + 2e- ---> Zn -0.76 then, you decide which one is reduced and which one is oxidized, the one with the more positive voltage is reduced, so in this case the copper is reduced. the overall reaction changes to Zn- ---> Zn2+ + 2e +0.76 Cu2+ + 2e- --> Cu +0.34 ---- Zn + Ni2+ ---> Ni + Zn2+ + 0.5 you add the voltages and get 0.5 as an experimental voltage

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