Chemistry · Redox Equilibrium · NEET
Write the reaction as a cell. The species that gets reduced (gains electrons) is the cathode; the species that gets oxidised (loses electrons) is the anode. Then compute E°cell = E°(cathode) − E°(anode) using standard reduction potentials from the data table. If E°cell is positive, the reaction is feasible (spontaneous). If it is negative, the reaction will not proceed on its own.
Positive E°cell means spontaneous. This is because ΔG° = −nFE°cell. When E°cell is positive, ΔG° becomes negative, and a negative ΔG° is the condition for a spontaneous reaction. So the sign flips: positive potential gives negative free energy gives a feasible reaction.
The half-cell with the higher (more positive) standard reduction potential acts as the cathode, because it has the stronger pull to be reduced. The half-cell with the lower (more negative) potential is forced to be the anode, where oxidation happens. If you assign them this way, E°cell always comes out positive for the real spontaneous direction.
Both values in the table are reduction potentials. But at the anode, oxidation actually happens, which is the reverse of reduction. Reversing a half-reaction flips the sign of its potential. So E°cell = E°(cathode, reduction) + E°(anode, oxidation) = E°(cathode) − E°(anode). The subtraction already handles the sign flip for you.
Not on its own. A negative E°cell means the reaction is non-spontaneous in that direction. You can only force it by supplying outside electrical energy, which is what happens in electrolysis (an electrolytic cell). For NEET feasibility questions, treat negative E°cell as 'the reaction does not occur' unless the question mentions electrolysis.
Reducing power is the tendency to lose electrons (to get oxidised). A more negative standard reduction potential means the species is more easily oxidised, so it is a stronger reducing agent. That is why the order of reducing power follows increasingly negative E° values: the most negative E° is the strongest reducing agent.
Given the half-cell reactions MnO4^- + 8H+ + 5e- -> Mn2+ + 4H2O with E°(MnO4^-/Mn2+) = +1.510 V, and 1/2 O2 + 2H+ + 2e- -> H2O with E°(O2/H2O) = +1.223 V. Will the permanganate ion MnO4^- liberate O2 from water in the presence of acid?
The standard electrode potentials of Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are -1.66 V, 0.80 V, -2.93 V and -0.74 V respectively. The correct decreasing order of reducing power of the metals is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A redox reaction is feasible if and only if E°cell is positive. Compute E°cell = E°(cathode) − E°(anode) and check the sign.
ΔG° = −nFE°cell, where n is the number of electrons transferred and F is the Faraday constant. Positive E°cell gives negative ΔG°, which means the reaction is spontaneous.
No. Standard potentials are intensive properties, so you do not multiply them by the number of electrons. Just subtract E°(anode) from E°(cathode) directly, regardless of how many electrons each half-reaction uses.
A higher (more positive) reduction potential means a stronger oxidising agent, because it has a greater tendency to be reduced. A more negative value means a stronger reducing agent.
Yes. For example, Zn (E° = −0.76 V) displaces Cu from Cu2+ (E° = +0.34 V) because E°cell = 0.34 − (−0.76) = +1.10 V, which is positive and therefore feasible.
