Chemistry · Redox Equilibrium · NEET
You subtract. The formula is E°cell = E°cathode - E°anode. A common wrong habit is adding the two numbers. Adding only works if you deliberately flipped one electrode into an oxidation potential first and changed its sign. To avoid mistakes, always keep both values as standard reduction potentials and just subtract.
Use standard reduction potentials for both electrodes. NEET tables and NCERT give reduction potentials. Oxidation potential is just the reduction value with the opposite sign. If you mix the two types, your answer will be wrong. Rule: keep everything as reduction potential, then E°cell = E°cathode - E°anode.
The electrode with the higher (more positive) standard reduction potential is the cathode (reduction happens there). The one with the lower value is the anode (oxidation happens there). Example: Cu (+0.34 V) and Zn (-0.76 V) - copper is the cathode, zinc is the anode. This choice makes E°cell come out positive, which is what a real galvanic cell needs.
No. Electrode potential is an intensive property - it does not depend on how many electrons or how you balance the equation. So even if one half-reaction has 5 electrons and the other has 2, you do NOT scale the E° value. You only balance electrons when finding delta-G or using the Nernst equation, not for E°cell itself.
A negative E°cell means the reaction as written is non-spontaneous - it will not run on its own (delta-G is positive). The reverse reaction is the spontaneous one. In NEET, a negative E°cell is your signal that the metal or ion you picked as the oxidiser is actually too weak, so the reaction does not proceed.
Half-cell reactions: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, E°(MnO4-/Mn2+) = -1.510 V (quoted for the oxidation direction); 1/2 O2 + 2H+ + 2e- -> H2O, E°(O2/H2O) = +1.223 V. Will MnO4- liberate O2 from water in acid?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E°cell = E°cathode - E°anode, using standard reduction potentials for both electrodes. It can also be written as E°(right) - E°(left) when the cell is drawn in standard convention (anode on the left, cathode on the right).
A positive E°cell means delta-G° = -nFE°cell is negative, so the reaction releases energy and runs spontaneously. That is how a galvanic (voltaic) cell produces current. A negative E°cell would need an outside power source (electrolytic cell).
No. Standard electrode potential is intensive, so it stays the same no matter how you multiply the half-reaction. You never multiply E° by n when finding E°cell.
Yes, but be consistent. If you use oxidation potentials, E°cell = E°ox(anode) + E°red(cathode). Most NEET students find it safer to keep only reduction potentials and use E°cathode - E°anode.
If E°cell comes out positive, the reaction as written is feasible (spontaneous). If negative, it is not - the reverse runs instead. This links directly to the feasibility topic tested every year in NEET.
