Electrochemical cells
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge.
For the cell
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag
The cell reaction is Ni(s) + 2Ag+ (aq) → Ni2+(aq) + 2Ag(s).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.
Which of the following is the anodic half cell reaction?
Ni(s) + 2Ag+ (aq) → Ni2+(aq) + 2Ag(s)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Ni(s) → Ni2+(aq) + 2e−
In every electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode. In a galvanic cell the anode is the negative electrode, because the electrons released by oxidation leave the cell from there.
In the cell reaction, nickel loses electrons and silver ions gain them:
\( \displaystyle \ce{Ni(s) -> Ni^2+(aq) + 2e^-} \quad \text{(oxidation, anode)} \)
\( \displaystyle \ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \quad \text{(reduction, cathode)} \)
This agrees with the cell notation Ni(s) | Ni2+(aq) || Ag+(aq) | Ag, in which the anode is written on the left.
Why the other options are incorrect
Option (2) Ni2+(aq) + 2e− → Ni(s) — Incorrect. This is a reduction, the reverse of what happens at the nickel electrode in this cell.
Option (3) Ag+(aq) + e− → Ag(s) — Incorrect. This reduction takes place at the cathode, not at the anode.
Option (4) Ag(s) → Ag+(aq) + e− — Incorrect. Silver is not oxidised in this cell; its ions are reduced.