How to Predict If a Redox Reaction Is Feasible Using E° Values
Chemistry · Electrochemistry · NEET
A redox reaction is feasible (it happens on its own) only when its cell potential E°cell is positive. You find E°cell = E°cathode − E°anode, using standard reduction potentials. Memory hook: "Positive cell, reaction will sell." If E°cell comes out negative, the reaction will NOT go forward.
Pick the reduced species as cathode (higher E°) and the oxidised species as anode (lower E°). Then E°cell = E°cathode − E°anode: positive means the redox reaction is feasible; negative means it will not happen.
Your doubts, answered
What is the simple rule to know if a redox reaction is feasible?
Calculate E°cell = E°cathode − E°anode using standard reduction potentials. If E°cell is POSITIVE, the reaction is feasible (spontaneous, it happens by itself). If E°cell is NEGATIVE, the reaction is not feasible and will not go forward. This one sign check is the whole trick NEET tests.
How do I pick which species is the cathode and which is the anode?
The species that gets REDUCED (gains electrons) is at the cathode. The species that gets OXIDISED (loses electrons) is at the anode. Look at the reaction you are given: whatever ends up reduced is the cathode, whatever ends up oxidised is the anode. Then plug both standard reduction potentials into E°cell = E°cathode − E°anode without flipping any signs.
Does a higher (more positive) reduction potential mean it gets reduced?
Yes. A more positive E° means that species really wants to gain electrons, so it acts as the oxidising agent and gets reduced (it is the cathode). The species with the lower (more negative) E° is forced to give up electrons, so it gets oxidised (the anode). So the higher-E° couple pulls electrons from the lower-E° couple.
Why does positive E°cell mean the reaction is spontaneous?
Because ΔG° = −nFE°cell. Here n is moles of electrons and F is the Faraday constant (both positive). When E°cell is positive, ΔG° becomes negative, and a negative ΔG° means the reaction is spontaneous. So positive E°cell and negative ΔG° both point to the same thing: the reaction happens on its own. This link is a common NEET question.
How do E° values tell me which metal displaces another from its salt?
A metal can displace (kick out) another metal from its salt solution only if the metal has a LOWER (more negative) E° than the metal ion in solution. For example, Zn (−0.76 V) can displace Cu (+0.34 V) from CuSO4 because putting Zn as anode and Cu²⁺ as cathode gives a positive E°cell. But Cu cannot displace Zn, because that reverse cell gives a negative E°cell.
What is disproportionation and how do I use E° to spot it?
Disproportionation is when the SAME element is both oxidised and reduced at once. Use the two E° steps around that species: treat the step to its right as the cathode (reduction) and the step to its left as the anode (oxidation). If E°cathode − E°anode is positive, that middle species disproportionates. NEET 2018 tested exactly this with bromine species.
⚠️ The NEET trap ✗ Flipping the sign of the anode's reduction potential before subtracting, e.g. writing E°cell = E°cathode + (−E°anode reversed) and doubling the sign, or thinking the more NEGATIVE E° species gets reduced. ✓ Use E°cell = E°cathode − E°anode with the standard REDUCTION potentials exactly as given (do not pre-flip signs). The more POSITIVE E° couple is reduced (cathode); the more negative one is oxidised (anode). Positive E°cell = feasible. 🧠 NEET loves the reverse trap: the species with the MOST POSITIVE E° is the strongest oxidising agent and gets reduced, NOT the reducing agent. Tl³⁺ (+1.26 V) is a strong oxidiser, never a reducing agent.
Real NEET questions
NEET 2018
For bromine species the standard reduction potentials of each successive step are: BrO4⁻ →(+1.82 V) BrO3⁻ →(+1.50 V) HBrO →(+1.595 V) Br2 →(+1.0652 V) Br⁻. The species undergoing disproportionation is:
A · Br2
B · BrO4⁻
C · BrO3⁻
D · HBrO ✓
Solution: A species disproportionates only if E°cell for (its reduction step) minus (its oxidation step) is positive. For HBrO: the step to its right (HBrO → Br2) is the cathode, E° = +1.595 V; the step to its left (BrO3⁻ → HBrO), reversed, is the anode, E° = +1.50 V. E°cell = 1.595 − 1.50 = +0.095 V, which is positive. So HBrO spontaneously disproportionates into Br2 and BrO3⁻. Answer: HBrO.
NEET 2019 Odisha
The E° values 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 the reducing power of the metals is:
A · Ag > Cr > Al > K
B · K > Al > Cr > Ag ✓
C · K > Al > Ag > Cr
D · Al > K > Ag > Cr
Solution: Reducing power is the tendency to get oxidised (give electrons). The MORE NEGATIVE the standard reduction potential, the stronger the reducing agent. Order of E° from lowest to highest: K (−2.93) < Al (−1.66) < Cr (−0.74) < Ag (+0.80). So reducing power decreases as K > Al > Cr > Ag. This is the same idea used to check feasibility: the most negative couple is oxidised at the anode.
NEET 2023 Phase 2
Given E°(Al⁺/Al)=+0.55 V, E°(Tl⁺/Tl)=−0.34 V, E°(Al³⁺/Al)=−1.66 V, E°(Tl³⁺/Tl)=+1.26 V. Identify the INCORRECT statement.
A · Al⁺ is unstable in solution
B · Tl can be more easily oxidised to Tl⁺ than to Tl³⁺
C · Al is more electropositive than Tl
D · Tl³⁺ is a better reducing agent than Tl⁺ ✓
Solution: A more positive reduction potential means a stronger OXIDISING agent and a WEAKER reducing agent. E°(Tl³⁺/Tl) = +1.26 V is highly positive, so Tl³⁺ is a strong oxidising agent, not a reducing agent. Therefore statement (D) is incorrect. The other statements check out: Al⁺ disproportionates (unstable), Tl → Tl⁺ is easier than Tl → Tl³⁺, and Al (very negative E°) is more electropositive than Tl.
Solved Electrochemistry NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is a redox reaction feasible when E°cell is positive or negative?
Positive. If E°cell = E°cathode − E°anode is positive, the reaction is feasible and spontaneous. A negative E°cell means the reaction will not happen on its own.
Do I need to multiply E° by the number of electrons?
No. E° is an intensive property, so you never multiply it by n when finding E°cell. You only use n later in ΔG° = −nFE°cell or in the Nernst equation.
How is E°cell linked to ΔG°?
By ΔG° = −nFE°cell. A positive E°cell gives a negative ΔG°, which means the reaction is spontaneous. Both are just two ways of saying the reaction is feasible.
Which agent has the most positive E°: oxidising or reducing?
The oxidising agent. The most positive standard reduction potential belongs to the strongest oxidising agent, because it grabs electrons easily. The most negative E° belongs to the strongest reducing agent.
Why does this matter for NEET?
Every year NEET asks you to compare E° values to decide feasibility, reducing/oxidising strength, metal displacement, or disproportionation. One sign mistake changes the answer, so the rule E°cell = E°cathode − E°anode and 'positive means feasible' is high-yield.
