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
- (a)C > B > A
- (b)A > C > B
- (c)B > C > A
- (d)None of the above
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.
- (a)C > B > 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)A > C > 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)None of the above — 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.
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.
Three habits make this question, and every question like it, mechanical. First, fix the convention in words rather than symbols: more negative means more readily oxidised means stronger reducing agent. Say it once as a sentence and the sign never gets lost. Second, sort the numbers on a line before ranking anything — placing −3, −1.2 and +0.5 on a number line makes it obvious which end is which, and it removes the commonest error, which is treating −3 as smaller in magnitude than −1.2 because the digit is smaller. Third, check the answer against a metal you know. Sodium and potassium are famously violent reducing agents and sit at the negative end; gold and silver are famously unreactive and sit at the positive end. If your ranking would make the positive-potential metal the best reducing agent, it has the series upside down. It is also worth noticing that this stem never names the metals — it gives only numbers, so no chemistry beyond the convention is being tested, and a candidate who knows the one rule cannot be defeated by unfamiliar elements.
- Standard electrode potentials are standard reduction potentials measured against the standard hydrogen electrode, which is assigned 0.00 V
- The more negative the standard electrode potential, the more readily the species is oxidised and the stronger a reducing agent it is; the more positive, the stronger an oxidising agent
- Lithium, at about −3.05 V, is the most negative common value and the strongest metallic reducing agent; fluorine, at +2.87 V, is the most positive and the strongest common oxidising agent
- Zinc is at −0.76 V and copper at +0.34 V, which is why zinc displaces copper from copper sulphate solution and copper cannot displace zinc
- For the values given here — B at −3 V, C at −1.2 V, A at +0.5 V — reducing power runs B > C > A and oxidising power runs exactly the other way, A > C > B
Reading down the column, the potential rises and the reducing power falls. The three highlighted rows give B > C > A — option (c).
- Ranking two negative potentials by digit size — −3 V is more negative than −1.2 V, so it means greater reducing power
- Reading a large positive potential as high reactivity; it means the species is easily reduced, which makes it a good oxidising agent, not a good reducing agent
- Forgetting that these are reduction potentials by convention; reversing the half-reaction reverses the sign
BPSC gives bare numbers and unnamed metals, so the item tests only whether the sign convention is understood and can be applied under time pressure. UPSC almost never asks the convention directly; it asks its consequences instead — which substance acts as the reducing agent in a blast furnace, what the electrodes of a lead-acid battery are made of — so the series has to be known as chemistry rather than as an ordering rule.
Consider the following statements: Coke is one of the materials of the charge added to blast furnace for the production of steel/iron. Its function is to I. Act as a reducing agent. II. Remove silica associated with the iron ore. III. Function as fuel, to supply heat. IV. Act as an oxidizing agent. Of these statements:
- (a) I and II are correct
- (b) II and IV are correct
- (c) I and III are correct
- (d) III and IV are correct
Answer(c) I and III are correct
The same idea applied to industry: something has to give up electrons to turn iron oxide into iron, and in a blast furnace that reducing agent is carbon. Both questions turn on knowing what 'reducing agent' means rather than on remembering a list.
Consider the following statements regarding a motor car battery: I. The voltage is usually 12 V. II. Electrolyte used is hydrochloric acid. III. Electrodes are lead and copper. IV. Capacity is expressed in ampere-hour. Which of the above statements are correct?
- (a) I and II
- (b) II and III
- (c) III and IV
- (d) I and IV
Answer(d) I and IV
The practical use of a potential difference between two electrodes. A lead-acid cell delivers about 2 volts per cell precisely because of the gap between the two half-cell potentials, which is the same quantity this question tabulates for three metals.
Which of the following pairs will give displacement reactions ?
- (a) NaCl solution and copper metal
- (b) AgNO3 solution and copper metal
- (c) MgCl2 solution and aluminium metal
- (d) FeSO4 solution and silver metal
Answer(b) AgNO3 solution and copper metal
The 70th CCE paper of December 2024 asked the applied form of exactly this rule. A displacement happens only when the free metal has the more negative potential — copper displaces silver, but silver cannot displace iron and aluminium cannot displace magnesium.
- practice — not a real PYQ
A metal X has a standard electrode potential of −0.76 V and a metal Y of +0.34 V. Which of the following will happen ?
- (a)Y will displace X from a solution of X's salt
- (b)X will displace Y from a solution of Y's salt
- (c)Neither metal will displace the other
- (d)Both displacements will occur
Answer(b) X will displace Y from a solution of Y's salt — the more negative potential is oxidised more readily, which is why zinc at −0.76 V displaces copper at +0.34 V from copper sulphate solution.
- practice — not a real PYQ
The standard hydrogen electrode is assigned a standard electrode potential of
- (a)+1.00 V
- (b)0.00 V
- (c)−1.00 V
- (d)+0.34 V
Answer(b) 0.00 V — it is the arbitrary reference point of the electrochemical series; +0.34 V belongs to the copper half-cell.