Skip to content
Question 115 of 115

Q.Consider the following reaction :
Cu(s)+2Ag+(aq)→2Ag(s)+Cu2+(aq)Cu(s) + 2Ag^+(aq) \rightarrow 2Ag(s) + Cu^{2+}(aq)

(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current.
(iii) Write the half-cell reactions taking place at cathode and anode.
Chandigarh CbseCBSE Class XII Board 2018Subjective· 3mImportance★★★★★
100% · 115/115 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Cell: Cu ∣ Cu2+ ∣∣ Ag+ ∣ AgCu\,|\,Cu^{2+}\,||\,Ag^{+}\,|\,Ag; current flows Ag→Cu externally (electrons Cu→Ag); cathode Ag++e−→AgAg^{+}+e^{-}\rightarrow Ag, anode Cu→Cu2++2e−Cu\rightarrow Cu^{2+}+2e^{-}.

Concept. Constructing a galvanic (Daniell-type) cell from a redox reaction — CBSE Class-12 electrochemistry.

Why. In Cu+2Ag+→2Ag+Cu2+Cu + 2Ag^+ \rightarrow 2Ag + Cu^{2+}, copper loses electrons (oxidation) and silver ions gain them (reduction). Oxidation occurs at the anode (left), reduction at the cathode (right).

  1. Cell notation. Cu(s) ∣ Cu2+(aq) ∣∣ Ag+(aq) ∣ Ag(s)Cu(s)\ |\ Cu^{2+}(aq)\ ||\ Ag^{+}(aq)\ |\ Ag(s)
  2. Direction of current. Electrons travel through the wire from the Cu anode to the Ag cathode; conventional current therefore flows in the opposite sense — from the silver electrode (cathode, +) to the copper electrode (anode, –) in the external circuit. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.