A voltaic cell is set-up with a silver electrode in a silver nitrate solution an
ID: 511512 • Letter: A
Question
A voltaic cell is set-up with a silver electrode in a silver nitrate solution and a nickel electrode in a nickel (II) nitrate solution.
Use the reduction potentials table in the notes/book.
What is the reaction occuring at the anode?
What is the reaction occuring at the cathode?
What is the potential of the cell, E°cell?
Ag(s) Ag+(aq) + e
E°cell = +0.54 V
E°cell = 1.06 V
E°cell = +1.06 V
Ag+(aq) + e Ag(s)
E°cell = 0.54 V
Ni2+(aq) + 2e Ni(s)
Ni(s) Ni2+(aq) + 2e
- A. B. C. D. E. F. G. H.What is the reaction occuring at the anode?
- A. B. C. D. E. F. G. H.What is the reaction occuring at the cathode?
- A. B. C. D. E. F. G. H.What is the potential of the cell, E°cell?
A.Ag(s) Ag+(aq) + e
B.E°cell = +0.54 V
C.E°cell = 1.06 V
D.E°cell = +1.06 V
E.Ag+(aq) + e Ag(s)
F.E°cell = 0.54 V
G.Ni2+(aq) + 2e Ni(s)
H.Ni(s) Ni2+(aq) + 2e
Explanation / Answer
Half reactions are:
2Ag+(aq) + 2e -----> 2Ag(s) Reduction, at cathode, oxidation number of Ag changes from +1 to 0.
Ni(s) Ni2+(aq) + 2e Oxidation, at anode, oxidation number of Ni changes from 0 to +2.
complete ionic reaction is:
Ni(s) + 2Ag+(aq) —> Ni2+(aq) + 2Ag(s)
Ecell = Ecathode - Eanode
= (-0.80V) - (-0.26)
= -0.54
Cell formed is: Ni(s) | Ni2+(aq) || Ag+ (aq) | Ag(s)
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