Which of the following is the balanced cell reaction of the galvanic…

The correct answer is Option (3) → \( \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \)

• In the notation of a galvanic cell the anode is written on the left and the cathode on the right. So the copper electrode is the anode, where oxidation takes place, and the silver electrode is the cathode, where reduction takes place.

      Anode (oxidation): \( \displaystyle \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \)

      Cathode (reduction): \( \displaystyle \ce{2Ag+(aq) + 2e^- -> 2Ag(s)} \)

• Each Ag⁺ ion takes only one electron, so the silver half reaction is multiplied by two to use the two electrons given by one copper atom. Adding the two half reactions gives the cell reaction:

      \( \displaystyle \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \)

Why the other options are incorrect

Option (1) \( \ce{Cu(s) + Ag+(aq) -> Cu^2+(aq) + Ag(s)} \) — Incorrect. The direction is right, but the equation is not balanced for charge: two electrons are lost by copper and only one is taken by one Ag⁺ ion.

Option (2) \( \ce{Cu^2+(aq) + 2Ag(s) -> Cu(s) + 2Ag+(aq)} \) — Incorrect. This is the reverse reaction. It would make copper the cathode and silver the anode, against the way the cell is written.

Option (4) \( \ce{Cu^2+(aq) + Ag(s) -> Cu(s) + Ag+(aq)} \) — Incorrect. It is the reverse reaction and is also not balanced for charge.

NCERT (2026-27 edition):
2.2 Galvanic Cells – “…It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell.”

“Cell reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2 Ag(s) (2.4)

Half-cell reactions:
Cathode (reduction): 2Ag⁺(aq) + 2e^– → 2Ag(s)   (2.5)
Anode (oxidation): Cu(s) → Cu²⁺(aq) + 2e^–   (2.6)

It can be seen that … silver electrode acts as a cathode and copper electrode acts as an anode. The cell can be represented as: Cu(s)|Cu²⁺(aq)||Ag⁺(aq)|Ag(s) …”