Question
The photo with the graphic is question 6. doesnt need to be answred. just need the...

![At what concentration of Zn2+ will the situation in question 6 become nonspontaneous if [Cu2+] = 0.1M? B IV AA- IE: - DO VE 1](http://img.homeworklib.com/questions/a7ae4770-d2ec-11eb-9d16-1b6ad9c235a7.png?x-oss-process=image/resize,w_560)
il Sprint 15%D 8:22 PM 7 of 17 Done Using line notation, write down what the following picture is describing Anode Cathode bridge HTML Editore BIVA-AI EI 3 1 X X, 5. OD 2N VIVO T 12pt MacBook Pro
At what concentration of Zn2+ will the situation in question 6 become nonspontaneous if [Cu2+] = 0.1M? B IV AA- IE: - DO VE 16V HTML Edit x x 12pt MacBook Pro onde
Answers
Anode : Zn(s) = Zn2+(aq) + 2e E° = 0.76
Cathode: Cu2+(aq) +2e = Cu(s) E° = 0.34
E°cell = 0.76 + 0.34 = 1.10 V
Line notation--
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Cell become non spontaneous when E<0
So, Ecell = E°cell - (0.059/n)log(Zn2+/Cu2+)
= Ecell = 1.10 - (0.059/2)log(Zn2+/0.1) < 0
=> [Zn2+] > 1.94*10^36 M
So, concentration of Zn2+ should be greater than 1.94*10^36 to make the cell reaction non spontaneous.
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