Standard electrode potential of three metals A, B and C are respectively 0.5 V, –3 V and –1.2

V. Reducing power of these metals would be

Question

Standard electrode potential of three metals A, B and C are respectively 0.5 V, –3 V and –1.2 V. Reducing power of these metals would be

UnlockIAS · Accepted Answer

Correct — C, B > C > A. The whole question rests on one relationship, and it runs the opposite way to the numbers. A standard electrode potential is by convention a standard reduction potential: it measures how readily a species accepts electrons and is reduced, against the standard hydrogen electrode, which is assigned exactly 0.00 volts. A large positive value therefore means the species is easily reduced and is a good oxidising agent. A large negative value means the reverse — the metal resists being reduced and instead gives up electrons readily, which is precisely what being a reducing agent means. So reducing power increases as the standard electrode potential becomes more negative. Rank the three by that rule. B stands at −3 V, the most negative of the three and close to lithium's −3.05 V, the most negative value in the whole table and the reason lithium is the strongest common reducing agent among metals. C stands at −1.2 V, still negative and therefore still willing to be oxidised, but far less eagerly than B. A stands at +0.5 V, a positive value: A is the one metal of the three that would rather be reduced than oxidised, so it is the weakest reducing agent and, of the three, the best oxidising agent. Reducing power therefore runs B > C > A, which is option (c). The physical consequence is worth carrying with the rule: a metal with a more negative potential will displace a metal with a less negative potential from a solution of its salt, which is why zinc at −0.76 V displaces copper at +0.34 V from copper sulphate solution and copper does not displace zinc. Option (d) fails on the simplest ground available — (c) states the correct order, so nothing is left over.

Why the other options are wrong — (a) Puts C ahead of B, which reverses the two negative values. At −3 V, B is more negative than C at −1.2 V, so B parts with electrons more readily and is the stronger reducing agent. The option is right about A being weakest, so it catches a candidate who has the rule but compares two negative numbers carelessly — −1.2 is larger than −3, not smaller.  (b) Exactly inverts the correct order, which is what happens if reducing power is read as increasing with the electrode potential rather than decreasing. This is the order of oxidising power, not reducing power: A at +0.5 V is indeed the best oxidising agent of the three. Ranking by the arithmetic size of the number without attending to the sign convention produces this answer.  (d) Only earns the mark if none of the three orders given is correct. Option (c) is correct, so this fails. A None-of-the-above option in a ranking question is worth a glance only when the data contain a tie or an impossible value, and there is neither here — the three potentials are distinct and all three orders offered are genuine permutations.

Concept: Every half-reaction has a standard electrode potential, measured under standard conditions against the standard hydrogen electrode, which is defined as zero. Arranging all the half-reactions by that value produces the electrochemical series, and the series is a single ordered list that answers a whole family of questions. At the negative end sit the alkali and alkaline-earth metals — lithium at about −3.05 V, then potassium, calcium, sodium, magnesium, aluminium, zinc at −0.76 V, iron — all of them ready to lose electrons and therefore reducing agents. Hydrogen sits at zero by definition. At the positive end sit copper at +0.34 V, silver, and finally fluorine at +2.87 V, the strongest common oxidising agent. Reading the series downwards from the negative end gives the reactivity series of metals that school chemistry teaches; reading which species is above which tells you the direction of every displacement reaction; and taking the difference between two potentials gives the electromotive force of the cell you would build from them, positive for a spontaneous cell. One list, several uses, which is why the sign convention is worth internalising rather than memorising case by case.

State PCS 2025 Q145 — Science & Technology PYQ with Answer

State PCS 2025 Q145 — Science & Technology PYQ with Answer