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In this problem you are asked to calculate Ecell, so you use Ecell = Ecathode - Eanode. E of the anode is given as -1.33V, so I don't understand why you make it positive, and then subtract it from the Ecathode. Can someone explain?
Ecell = Ecathode - Eanode is the equation you use when Ecathode and Eanode are both given as reduction potentials (meaning when both reactions are written as reduction reactions with the elections on the reactant side). However, since the second equation 2Cr3+ (aq) + 7H2O (l) --> Cr2O7 2- (aq) +14H+ (aq) +6e- , is flipped to be the oxidation reaction, they also flipped the sign of the reduction potential. So the reduction potential for the second equation, when written as > Cr2O7 2- (aq) +14H+ (aq) +6e- --> 2Cr3+ (aq) + 7H2O (l) , is positive 1.33V. Therefore Ecell = Ecathode - Eanode is 1.64-1.33=0.31V.
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