Please help with G.Chem question !!

[avian777]

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Calculating standard potential for this rx
Standard Red Potential (E)
Ce4+(aq) + e- ---> Ce3+ (aq) + 1.61 V
Br2 (l) + 2e- ---> 2 Br- (aq) + 1.06 V

I've been reasoning that since the first half rx has a higher reduction potential so Cerium 4+ is more likely to be reduced, and the second rx is the oxidation half and I reverse the second reaction so the potential will have negative sign and add them up to get cell potential of +.55 V

However, the answer from ACS G.chem is -.55V and they said you should reverse the first reaction. this is very confusing for me because i always get right in problems like this in other books by my way, McMurry textbook also seems agree with my thinking and on website they said " reversing which reaction will yield most positve standard reduction potential" Is there something that i'm missing ?
by the way, if the answer is -.55 V as in ACS book, then this rx is nonspontaneous, and it's spontaneous the other way around. Can someone help me with this??? thanks

 

[UCB05]

You didn't put down what the reaction in question is. (or the problem didnt say, in which case it's a bad question)
The standard potential for the reaction Br2 + Ce3+ --> Br- + Ce4+ is -.55V
The standard potential for the reaction Ce4+ + Br- --> Br2 + Ce3+ is +.55V

the first is nonspontaneous, the second is spontaneous. They are well within their rights to ask you what the standard potential for a nonspontaneous reaction is, and if they did, the answer is -.55V

 

[avian777]

I get it now, thanks UCB05