Skip to content
Question 40 of 52

Q.Calculate the equilibrium constant at 25 degC of the following equation of a Daniell cell: Zn(s) + Cu2+(aq) <=> Zn2+(aq) + Cu(s). [Given: E-standard(Zn2+/Zn) = -0.76 V and E-standard(Cu2+/Cu) = +0.34 V, R = 8.314 JK-1mol-1] OR Find the relation between resistance, specific conductance and cell constant of the electrolytic solution of the conductivity Daniell cell. Establish the relation between molar conductance and specific conductance of a solute. (1+2)

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 3mImportance★★★★★
77% · 40/52 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 →

Compute E°cell from the two standard electrode potentials, then use log⁡K=nE∘0.0591\log K = \dfrac{nE^\circ}{0.0591} at 298 K (from ΔG∘=−nFE∘=−RTln⁡K\Delta G^\circ=-nFE^\circ=-RT\ln K).

Step 1 — Identify cathode/anode and E°cell:

Cell reaction: Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)Zn(s)+Cu^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Cu(s)

Cu2+/Cu is reduced (cathode), Zn is oxidised (anode).

Ecell∘=Ecathode∘−Eanode∘=0.34−(−0.76)=1.10 VE^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-0.76)=1.10\text{ V}

Step 2 — Number of electrons transferred: n=2n=2 (Zn → Zn2+ + 2e-; Cu2+ + 2e- → Cu).

Step 3 — Relate E°cell to K:

From ΔG∘=−nFE∘=−RTln⁡K\Delta G^\circ=-nFE^\circ=-RT\ln K:

ln⁡K=nFE∘RT=2×96500×1.108.314×298=2123002477.6≈85.7\ln K=\frac{nFE^\circ}{RT}=\frac{2\times96500\times1.10}{8.314\times298}=\frac{212300}{2477.6}\approx85.7 …

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.