Start by finding the standard reduction potential for the Ag^+ and Ni^(2+) ion. Normally, the values are listed at the back of most chemistry textbooks.
Ag^+(aq) + 1e^(-) -> Ag(s)" " E^o= 0.80 \ V
Ni^(2+)(aq) + 2e^(-) -> Ni(s)" " E^o= -0.23 \ V
In the galvanic cell, the reaction is spontaneous and for a spontaneous reaction E_(cell)^o must be a positive quantity.
E_(cell)^o = E_(Anode)^o + E_(cathode)^o
Manipulate the two equations so that E_(cell)^o is positive. Note that the anode is the site of oxidation (where electrons are lost) and the cathode (where electron are gained) is the site for reduction.
Ag^+(aq) + color(red)(1e^(-)) -> Ag(s)" " E^o= 0.80 \ V
Ni(s) -> Ni^(2+)(aq) + color(red)(2e^(-)) " " E^o= 0.23 \ V
2xx{Ag^+(aq) + 1e^(-) -> Ag(s)}" " E^o= 0.80 \ V " " color(red)((Cathode))
Ni(s) -> Ni^(2+)(aq) + 2e^(-) " " E^o= 0.23 \ V " " color(red)((Anode)"
ul" "
2Ag^+(aq) + Ni(s) -> Ni^(2+)(aq) + 2Ag(s) " "E_(cell)^o = 1.03 \ V
Start with the anode components (site of oxidation) - the cathode components are listed to the right.
Ni(s) | Ni^(2+)(aq)\ || \ Ag^+(aq) | Ag(s)
The single vertical lines indicate the boundary (phase difference) between solid Ni and Ni^(2+) ions in the aqueous solution of the first compartment and between solid Ag and Ag^(+) ions present in the aqueous solution of the second compartments.
The double vertical lines refer to the salt bridge - note that the salt bridge must be an inert salt to both ions present in both compartments of the galvanic cell ...
