HW 14. 15 part a

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HW 14. 15 part a

Postby mayasinha1B » Mon Feb 19, 2018 8:49 pm

For homework problem 15 part a how do you get the half reactions displayed? Ag solid is not displayed in the skeleton reaction so how would you know to add this

Vasiliki G Dis1C
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Re: HW 14. 15 part a

Postby Vasiliki G Dis1C » Thu Feb 22, 2018 1:52 pm

For this kind of question, you want to find the reduction potentials of the different components of the reaction you are given. So you need the reduction potential of Ag+ which is Ag+ + e- -> Ag(s) and has an E°=0.80 V, and the reduction potential of AgBr which is AgBr(s) + e- -> Ag + Br- and has E°=0.07 V. Since 0.07V<0.80V, the AgBr will be oxidized and the Ag will be the cathode. Since we have determined that the AgBr will be the anode, we reverse this reaction to get Ag + Br- -> AgBr(s) + e-. Now, we combine both these reactions to get the overall cell reaction, which is Ag+ + Br- -> AgBr(s) and has an E°=0.80-0.07 = 0.73V.
The cell diagram to model this would be Ag(s)|AgBr(s)|Br-||Ag+|Ag(s)
Basically, you need the Ag solid when you look a the reduction potential of each part of the reaction. Hope this helps!

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