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^+???

Question

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

DrBob222 · Answer

Ecell = Eocell - (0.0592/n)log Q Proof your post. You show NiO2 as a solid yet you ask a question about NiO^2+

Anonymous · Answer

Sorry I meant Ni^2+.....

DrBob222 · Answer

OK. Then the above equation I wrote will do it. Q = (Ag^+)^2(Ni^2+)/(H^+)^4