JEE / JEE Main 2027 / Chemistry / Redox Reactions and Electrochemistry Section 3 of 5 40 min read Medium Practice — Set 1 Quick Recap — Cell Potential, Nernst & Faraday Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}Ecell∘=Ecathode∘−Eanode∘; Nernst (298 K): E=E∘−0.059nlogQE=E^\circ-\dfrac{0.059}{n}\log QE=E∘−n0.059logQ. At equilibrium E=0E=0E=0: E∘=0.059nlogKE^\circ=\dfrac{0.059}{n}\log KE∘=n0.059logK; ΔG∘=−nFE∘\Delta G^\circ=-nFE^\circΔG∘=−nFE∘. Faraday: m=MnFItm=\dfrac{M}{nF}Itm=nFMIt; 111 mol electrons =96500=96500=96500 C. Molar conductivity: Λm=κ×1000C\Lambda_m=\dfrac{\kappa\times1000}{C}Λm=Cκ×1000 (κ in S/cm, C in mol/L). Ready to test your knowledge? Take a quick interactive quiz on this topic — free, works without login. Start Quiz Next: Medium Practice — Set 2 ← Easy Practice — Set 2 Medium Practice — Set 2 →