Asked by Anonymous

The voltaic cell reaction NiO2(s)+4H^+(aq)+ 2Ag(s)-->Ni^2+(aq)+2H20(l)+ 2Ag^+ has a E cell= +2.48V. What will the cell potential be at a pH of 5.00 when the concentration of NiO^2+ and Ag^+???

I need to know how to set this type of problem up to solve it

Answers

Answered by DrBob222
Ecell = Eocell - (0.0592/n)log Q
Proof your post. You show NiO2 as a solid yet you ask a question about NiO^2+
Answered by Anonymous
Sorry I meant Ni^2+.....
Answered by DrBob222
OK. Then the above equation I wrote will do it. Q = (Ag^+)^2(Ni^2+)/(H^+)^4
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