Question
Calculate the cell potential for this voltaic cell at 298K if it is constructed using solutions of .25M Ni(NO3)2 and .01M Cu(NO3)2
Any explanation would be spectacular
Any explanation would be spectacular
Answers
DrBob222
Look up Eo for Ni^2+ + 2e ==> Ni(s) as a reduction potential.
Do the same for Cu^2+ + 2e ==> Cu(s)
Reverse the Ni equation, change sign of Eo for Ni and add Ni Eo(as an oxidation) to Cu as a reduction to arrive at Eocell and this equation.
Ni(s) + Cu^2+(0.01M) ==> Ni^2+(0.25M) + Cu(s)
Then
Ecell = Eocell - (0.0592/2)log Q
where Q = (Ni^2+)(Cu)/(Ni)(Cu^2+)
Do the same for Cu^2+ + 2e ==> Cu(s)
Reverse the Ni equation, change sign of Eo for Ni and add Ni Eo(as an oxidation) to Cu as a reduction to arrive at Eocell and this equation.
Ni(s) + Cu^2+(0.01M) ==> Ni^2+(0.25M) + Cu(s)
Then
Ecell = Eocell - (0.0592/2)log Q
where Q = (Ni^2+)(Cu)/(Ni)(Cu^2+)