Winter 2013 Q3B

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Nadine Schenker 3E
Posts: 7
Joined: Wed Sep 21, 2016 2:58 pm

Winter 2013 Q3B

Postby Nadine Schenker 3E » Fri Mar 17, 2017 8:01 pm

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?

Fengting Liang 1F
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Winter 2013 Q3B

Postby Fengting Liang 1F » Fri Mar 17, 2017 8:54 pm

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.

Cayla_Brooks_1I
Posts: 10
Joined: Mon Jan 26, 2015 2:17 pm

Re: Winter 2013 Q3B

Postby Cayla_Brooks_1I » Sun Mar 19, 2017 2:07 pm

I do not understand how to determine which is the anode and which is the cathode if both of the reactions appear to be being oxidized.


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