Skip to content
Exercise · Q2

Q.A galvanic cell is built by combining a silver electrode dipped in 1 M AgNO31\ \text{M}\ \text{AgNO}_3 with a copper electrode dipped in 1 M CuSO41\ \text{M}\ \text{CuSO}_4. Using the standard reduction potentials E∘(Ag+/Ag)=+0.80 VE^{\circ}(\text{Ag}^{+}/\text{Ag}) = +0.80\ \text{V} and E∘(Cu2+/Cu)=+0.34 VE^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}, identify the cathode, the anode, the oxidizing agent, and the reducing agent.

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
4% · 2/52 Questions
✓ Free question

In any galvanic cell, the electrode with the higher (more positive) standard reduction potential is reduced and therefore acts as the cathode, while the electrode with the lower standard reduction potential is oxidized and acts as the anode. Here E∘(Ag+/Ag)=+0.80 VE^{\circ}(\text{Ag}^{+}/\text{Ag}) = +0.80\ \text{V} is more positive than E∘(Cu2+/Cu)=+0.34 VE^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}, so silver ions are preferentially reduced: 2Ag+(aq)+2e−→2Ag(s)2\text{Ag}^{+}(aq) + 2e^{-} \to 2\text{Ag}(s) at the cathode, and copper metal is oxidized: Cu(s)→Cu2+(aq)+2e−\text{Cu}(s) \to \text{Cu}^{2+}(aq) + 2e^{-} at the anode. The overall spontaneous cell reaction is Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\text{Cu}(s) + 2\text{Ag}^{+}(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s). The species that is reduced (Ag+\text{Ag}^+) is, by definition, the oxidizing agent (it causes copper to be oxidized), and the species that is oxidized (Cu\text{Cu}) is the reducing agent. [!ANSWER] Cathode = silver electrode; anode = copper electrode; oxidizing agent = Ag+\text{Ag}^+; reducing agent = Cu\text{Cu}.

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.