Page 1 of 1

HW 14. 15 part a

Posted: Mon Feb 19, 2018 8:49 pm
by mayasinha1B
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

Re: HW 14. 15 part a

Posted: Thu Feb 22, 2018 1:52 pm
by Vasiliki G Dis1C
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!