Q.Given the standard electrode potentials:
K+/K=−2.93 V, Ag+/Ag=0.80 V, Hg2+/Hg=0.79 V,
Mg2+/Mg=−2.37 V, Cr3+/Cr=−0.74 V
Arrange these metals in their increasing order of reducing power.
Concept understanding — Standard Electrode Potentials
Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"
Dip a zinc rod into a zinc-salt solution and a tiny tug-of-war begins at the surface:
metal atoms tend to dissolve as ions (leaving electrons behind on the rod) while ions
from the solution tend to deposit as metal (consuming electrons). The rod ends up with
a characteristic electrical potential relative to the solution — the electrode potential. It is a direct measure of the tendency of that redox couple to gain or
lose electrons.
The Core Idea
Different couples pull electrons with very different strengths. Copper's ion grabs
them readily; zinc's barely wants them. Put a number on each couple and you can
predict, before mixing anything, who will oxidise whom.
Two conventions make the numbers comparable:
- Standard conditions. Every species at unit concentration (1 M), any gas at 1 atm, temperature 298 K. The potential measured then is the standard electrode potential, written E⊖.
- A common zero. Potentials can only be measured as differences, so one electrode is defined as the reference: the standard hydrogen electrode (SHE), 2H++2e−→H2, is fixed at exactly 0.00 V. Every E⊖ is the voltage of a couple measured against it.
By convention the values are tabulated for the reduction direction:
Oxidised form+ne−→Reduced formE⊖ (in volts, at 298 K)
Reading the Table
The standard-potential table (Table 7.1 in the Class 11 chapter) runs from
F2/F− at +2.87 V down to Li+/Li at −3.05 V.
Two rules unlock it:
- More positive E⊖ → stronger oxidising agent (the oxidised form is hungrier for electrons). F₂ tops the table; that is why fluorine oxidises almost everything.
- More negative E⊖ → stronger reducing agent (the reduced form gives electrons up most easily). Li, K, Ca, Na at the bottom are the great electron donors. A negative E⊖ means the couple is a stronger reducing agent than the H⁺/H₂ couple; a positive one, weaker.
Predicting Whether a Reaction Goes
For any proposed redox reaction, the species being reduced acts as the cathode couple
and the species being oxidised as the anode couple:
Ecell⊖=Ecathode⊖−Eanode⊖
A positive Ecell⊖ means the reaction is feasible
(spontaneous) under standard conditions; a negative one means the reverse reaction
is the spontaneous direction.
Worked feel: can Fe³⁺ oxidise iodide? E⊖(Fe3+/Fe2+)=+0.77 V is above E⊖(I2/I−)=+0.54 V, so
Ecell⊖=+0.23 V — yes. Can silver metal reduce Fe³⁺?
0.77−0.80=−0.03 V — no.
This is also the logic of the activity series: a metal displaces, from solution,
the ions of any metal whose couple sits above it (less negative / more positive) in
the table — Zn displaces Cu²⁺, Cu displaces Ag⁺, never the reverse.
Common Mistakes
- Flipping the sign when a half reaction is reversed and then double-counting. Use Ecell⊖=Ecathode⊖−Eanode⊖ with BOTH values as tabulated (reduction) potentials — the subtraction already handles the reversal.
- Multiplying E⊖ by stoichiometric coefficients. Potentials are intensive: doubling a half reaction does not double its E⊖.
- Reading "negative" as "impossible". A negative E⊖ only ranks the couple as a stronger reducing agent than H⁺/H₂ — zinc's −0.76 V is exactly why zinc is so good at reducing other ions.
Exam Relevance
Standard-potential questions are staples: pick the strongest oxidising/reducing agent
from given E⊖ values, decide whether a pair reacts (CBSE Class 11 Exercise
7.26; Exemplar Q2–Q4, Q16, Q34), order metals by reducing power, or justify a
displacement series. In Class 12 the same idea grows into full electrochemistry — the
Nernst equation extends E⊖ to non-standard concentrations, and cell EMF
connects to thermodynamics via ΔG=−nFE.
Why this formula?
Galvanic Corrosion: Why the Key Formulas Hold
Galvanic corrosion occurs when two dissimilar metals are electrically connected in the presence of an electrolyte. The key formula that governs this is the mixed potential theory, which leads to the galvanic current and corrosion rate expressions.
Let's build the reasoning step-by-step.
1. The Core Idea: Two Electrodes, One Circuit
When metals M₁ (more active, e.g., zinc) and M₂ (more noble, e.g., copper) are connected:
- M₁ acts as the anode — it oxidizes (corrodes):
M1→M1n++ne−
- M₂ acts as the cathode — it reduces something (e.g., oxygen or H⁺):
O2+2H2O+4e−→4OH−(in neutral/alkaline)
or
2H++2e−→H2(in acidic)
The two metals are electrically connected (via a wire or direct contact), and the electrolyte completes the circuit. Electrons flow from M₁ to M₂.
2. The Mixed Potential: Why It Exists
Each metal, when alone in the electrolyte, has its own open-circuit potential (OCP) — the equilibrium potential for its half-reaction. For M₁, it's Ecorr,1; for M₂, it's Ecorr,2.
When connected, the system cannot stay at two different potentials. The entire metal couple must reach a single potential — the mixed potential Emix.
- Emix lies between Ecorr,1 and Ecorr,2.
- At Emix, the total anodic current from M₁ equals the total cathodic current from M₂ (charge conservation):
Ianode=Icathode
This is the fundamental equation of galvanic corrosion.
3. Deriving the Galvanic Current
Assume each electrode follows Butler-Volmer kinetics (for activation-controlled reactions). For the anode (M₁), the anodic current density ia at potential E is:
ia=i0,1exp(RTαaF(E−E0,1))
For the cathode (M₂), the cathodic current density ic is:
ic=i0,2exp(−RTαcF(E−E0,2))
Where:
- i0,1,i0,2 = exchange current densities
- αa,αc = transfer coefficients (typically ~0.5)
- F = Faraday constant
- R = gas constant
- T = temperature
- E0,1,E0,2 = standard reduction potentials
At the mixed potential Emix:
Igalvanic=A1⋅ia(Emix)=A2⋅ic(Emix)
Where A1 and A2 are the surface areas of the anode and cathode.
Why this holds: The net current from the anode must exactly balance the net current consumed at the cathode — otherwise, charge would accumulate, which is impossible in a steady-state circuit.
4. The Corrosion Rate Formula
The corrosion rate (mass loss per time) of the anode is given by Faraday's law:
Corrosion rate=n⋅F⋅ρIgalvanic⋅M
Where:
- M = molar mass of the anode metal
- n = number of electrons transferred per atom
- ρ = density of the metal
- F = Faraday constant (96,485 C/mol)
Why this holds: Each mole of metal oxidized releases n moles of electrons. The total charge passed Q=Igalvanic⋅t corresponds to moles of metal lost:
moles lost=nFQ=nFIgalvanic⋅t
Multiply by M/ρ to get volume or thickness loss.
5. The Area Effect: Why It Matters
From the mixed potential equation:
A1⋅ia(Emix)=A2⋅ic(Emix)
If the cathode area A2 is large relative to the anode area A1, then ia(Emix) must be large to balance the current. This means:
- Small anode + large cathode → severe galvanic corrosion (high current density on the anode).
- Large anode + small cathode → mild corrosion.
Why this holds: The current density on the anode is ia=Igalvanic/A1. For a fixed Igalvanic, smaller A1 gives higher ia, which accelerates corrosion.
6. The Driving Force: Potential Difference
The driving force for galvanic corrosion is the difference in open-circuit potentials:
ΔE=Ecorr,2−Ecorr,1
A larger ΔE generally leads to a larger Igalvanic, but the exact relationship depends on the polarization behavior (Tafel slopes) of both electrodes.
Why this holds: The mixed potential Emix is determined by the intersection of the anodic and cathodic polarization curves. A larger separation between the two curves shifts the intersection to a higher current.
Summary of Key Takeaways
| Concept | Formula | Why It Holds |
|---|---|---|
| Mixed potential | Ianode=Icathode | Charge conservation in a closed circuit |
| Galvanic current | A1ia(Emix)=A2ic(Emix) | Butler-Volmer kinetics + area balance |
| Corrosion rate | nFρIgalvanicM | Faraday's law of electrolysis |
| Area effect | Small anode → high ia | Current density inversely proportional to area |
| Driving force | ΔE=Ecorr,2−Ecorr,1 | Larger potential difference → larger current (generally) |
Exam tip: Always start with the mixed potential condition — it's the foundation. Then apply Faraday's law for the rate. Never forget the area ratio — it's the most common trick in exam problems.
Concept: Standard Electrode Potentials — A more negative reduction potential means the metal is a stronger reducing agent (it loses electrons more readily).
Reasoning:
- Reducing power is the ability to donate electrons. The more negative the standard reduction potential (E∘), the stronger the reducing agent.
- Arrange the given E∘ values from most negative to most positive: K+/K=−2.93 V (most negative → strongest reducing agent) Mg2+/Mg=−2.37 V Cr3+/Cr=−0.74 V Hg2+/Hg=0.79 V Ag+/Ag=0.80 V (most positive → weakest reducing agent)
- The increasing order of reducing power is the reverse of this list: weakest to strongest.
The increasing order of reducing power is: Ag<Hg<Cr<Mg<K.
Reducing power increases as the standard electrode potential becomes more negative. The order of increasing reducing power is: Ag < Hg < Cr < Mg < K.
The reducing power of a metal is its ability to lose electrons and get oxidised. In electrochemistry, this is directly linked to the standard electrode potential (E∘) of the metal/metal-ion half-cell.
A more negative E∘ means the metal is more easily oxidised — it is a stronger reducing agent. A more positive E∘ means the metal is harder to oxidise — it is a weaker reducing agent (and its ions are stronger oxidising agents).
So, to arrange metals in increasing order of reducing power, we need to go from the most positive E∘ (weakest reducing agent) to the most negative E∘ (strongest reducing agent).
Let's list the given potentials:
| Metal | E∘ (V) |
|---|---|
| Ag | +0.80 |
| Hg | +0.79 |
| Cr | -0.74 |
| Mg | -2.37 |
| K | -2.93 |
-
Identify the weakest reducing agent. The most positive potential is Ag+/Ag at +0.80 V. Silver is the least willing to lose electrons, so it has the least reducing power. It comes first.
-
Next comes mercury. Hg2+/Hg is +0.79 V, very close to silver but slightly less positive. So Hg is a slightly stronger reducing agent than Ag, but still much weaker than the others. It comes second.
-
Now we cross into negative potentials. Chromium has E∘=−0.74 V. This is significantly more negative than Ag or Hg, meaning Cr is a much stronger reducing agent. It comes third.
-
Magnesium follows. Mg2+/Mg at −2.37 V is more negative than Cr, so Mg is a stronger reducing agent than Cr. It comes fourth.
-
Potassium is the strongest. K+/K at −2.93 V is the most negative potential given. Potassium is the most powerful reducing agent in this list. It comes last.
A common mistake is to confuse reducing power with the tendency to get reduced. Remember: more negative E∘ = stronger reducing agent. If you accidentally arrange by increasing tendency to get reduced (i.e., by increasing E∘), you would get the exact opposite order.
You can think of the electrochemical series as a ladder: metals at the top (like Li, K, Ca) have very negative E∘ and are strong reducing agents. Metals at the bottom (like Au, Pt, Ag) have very positive E∘ and are weak reducing agents. Here, K is near the top, Ag is near the bottom.
The increasing order of reducing power is Ag < Hg < Cr < Mg < K.
Method: Using Standard Electrode Potentials to Compare Reducing Power
Method Name: The More Negative E∘, the Stronger the Reducing Agent
Concept (Why this works)
Reducing power means the ability to lose electrons (get oxidized).
A more negative standard reduction potential (E∘) means the metal is more difficult to reduce — which means it is easier to oxidize (i.e., it is a stronger reducing agent).
Steps
-
List the given E∘ values (all are reduction potentials):
- K+/K: −2.93 V
- Mg2+/Mg: −2.37 V
- Cr3+/Cr: −0.74 V
- Hg2+/Hg: +0.79 V
- Ag+/Ag: +0.80 V
-
Arrange in increasing order of E∘ (most negative → most positive):
- −2.93 V (K)
- −2.37 V (Mg)
- −0.74 V (Cr)
- +0.79 V (Hg)
- +0.80 V (Ag)
-
Apply the rule:
More negative E∘ = stronger reducing agent.
So increasing reducing power means going from weakest (most positive) to strongest (most negative).
-
Final order (increasing reducing power):
Ag < Hg < Cr < Mg < K
Key Result
Increasing reducing power:
Ag<Hg<Cr<Mg<K
Quick Check
- Ag (+0.80 V) is the weakest reducing agent — it prefers to stay as Ag+ rather than lose electrons.
- K (−2.93 V) is the strongest — it readily loses electrons to become K+.
Mistake 1: Confusing Reducing Power with Oxidising Power
The error:
Students often think: "Higher (more positive) electrode potential = stronger reducing agent."
This is wrong. Reducing power is the ability to lose electrons (get oxidised). A more negative E∘ means the metal is more eager to give away electrons.
How to avoid:
Remember the mnemonic:
More negative = more reactive (as a reducing agent).
- K+/K=−2.93 V -> very strong reducing agent
- Ag+/Ag=+0.80 V -> very weak reducing agent
Correct order (increasing reducing power):
Ag<Hg<Cr<Mg<K
Mistake 2: Forgetting the Sign Convention
The error:
Some students rank by absolute value (ignoring the negative sign), e.g., putting K and Mg in the middle.
How to avoid:
Always compare the actual signed values on a number line:
-3.0 V -2.0 V -1.0 V 0 V +1.0 V
K Mg Cr Hg Ag
(most negative = strongest reducing agent)
Increasing reducing power = from most positive to most negative E∘.
Mistake 3: Mixing Up the Order (Ascending vs Descending)
The error:
The question asks for increasing order of reducing power. Students often write the decreasing order (strongest first).
How to avoid:
- Increasing = weakest first -> strongest last
- Decreasing = strongest first -> weakest last
For this data:
- Increasing reducing power: Ag<Hg<Cr<Mg<K
- Decreasing reducing power: K>Mg>Cr>Hg>Ag
Mistake 4: Not Writing the Metal Symbols Correctly
The error:
Writing K+, Ag+ etc. instead of the metal (K, Ag). The question asks for metals, not ions.
How to avoid:
The reducing agent is the metal in its elemental form (M), not the ion (Mn+). Always list the neutral metal.
Final Correct Answer
Increasing order of reducing power:
Ag<Hg<Cr<Mg<K
Reasoning:
- Ag+/Ag=+0.80 V -> weakest reducing agent
- K+/K=−2.93 V -> strongest reducing agent
Quick Checklist to Avoid These Mistakes
| Mistake | Fix |
|---|---|
| Confusing reducing/oxidising power | More negative E∘ = stronger reducing agent |
| Ignoring negative signs | Plot on a number line |
| Reversing order | Read "increasing" carefully |
| Writing ions instead of metals | List only the neutral metal symbols |
- KEAM 2026Set eng-2026-04184 marksMCQQ.Which of the following 3d transition metal has the positive standard electrode potential (E0)? (A) Ni (B) Cu (C) V (D) Mn (E) Cr
›Reveal solutionSolution
Copper is the exception in the 3d series with a positive E0.
Standard reduction potentials (M2+/M) for the 3d series are generally negative because of high atomization plus ionization energies, e.g. V(−1.18), Cr(−0.91), Mn(−1.18), Ni(−0.25) V.
Copper is unique: its high enthalpy of atomization and ionization is not offset by hydration, giving ECu2+/Cu0=+0.34V (positive). This is why Cu does not liberate H2 from dilute acids.
✓Final answerThe correct option is (B).
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.Which of the following half-cell reaction has the most negative standard electrode potential? (A) Li(aq)++e−→Li(s) (B) F2(g)+2e−→2F(aq)− (C) Na(aq)++e−→Na(s) (D) I2(aq)+2e−→2I(aq)− (E) Cu(aq)++e−→Cu(s)
›Reveal solutionSolution
Standard reduction potentials: Li+/Li = -3.04 V is the most negative among the given couples.
Standard electrode (reduction) potentials:
- Li++e−→Li: -3.04 V
- Na++e−→Na: -2.71 V
- Cu++e−→Cu: +0.52 V
- I2+2e−→2I−: +0.54 V
- F2+2e−→2F−: +2.87 V (most positive)
The most negative is lithium.
✓Final answerThe correct option is (A). Li+/Li at -3.04 V is the strongest reducing couple.
- KEAM 2025Set eng-2025-04274 marksMCQQ.For which of the following electrode reactions the standard electrode potential is the highest at 298 K? The ions are present in aqueous solution. (A) Co3++e−→Co2+ (B) Cl2(g)+2e−→2Cl− (C) MnO2(s)+4H++2e−→Mn2++2H2O (D) F2(g)+2e−→2F− (E) AgCl(s)+e−→Ag(s)+Cl−
›Reveal solutionSolution
Fluorine is the strongest oxidising agent, so the F2/F− couple has the highest standard reduction potential (+2.87V).
Reasoning
Standard reduction potentials (aqueous, 298 K):
- (A) Co3+/Co2+: E0=+1.81V
- (B) Cl2/Cl−: E0=+1.36V
- (C) MnO2/Mn2+: E0=+1.23V
- (D) F2/F−: E0=+2.87V
- (E) AgCl/Ag: E0=+0.22V
The most positive value corresponds to the strongest oxidising agent. Fluorine's +2.87V is the highest, reflecting its very high electronegativity, low bond-dissociation energy and high hydration energy of F−.
✓Final answerThe correct option is (D).
- KEAM 2025Set eng-2025-04274 marksMCQQ.Acidified potassium dichromate cannot oxidize (A) Iodides to iodine (B) Iron (II) salt to iron (III) salt (C) Tin (II) salt to tin (IV) salt (D) H2S to sulphur (E) Fluoride to fluorine
›Reveal solutionSolution
The dichromate couple's potential (+1.33V) is far below fluorine's (+2.87V). It can oxidise I−, Fe2+, Sn2+ and H2S, but not F−.
Reasoning
Acidified potassium dichromate is a strong oxidiser:
Cr2O72−+14H++6e−→2Cr3++7H2O,E0=+1.33V
An oxidant can oxidise a species only if the reductant's own couple has a lower reduction potential. Comparing:
- I2/I−: +0.54V — oxidised (A) ✓
- Fe3+/Fe2+: +0.77V — oxidised (B) ✓
- Sn4+/Sn2+: +0.15V — oxidised (C) ✓
- S/H2S: +0.14V — oxidised (D) ✓
- F2/F−: +2.87V — NOT oxidised (E), since +2.87>+1.33
Fluorine is the strongest oxidant of all; nothing weaker than it (including dichromate) can convert F− to F2.
✓Final answerThe correct option is (E).
- KEAM 2024Set eng-2024-06064 marksMCQQ.The 3d block metal having positive standard electrode potential (M2+/M) is (A) Titanium (B) Vanadium (C) Iron (D) Copper (E) Chromium
›Reveal solutionSolution
Copper is exceptional in the first transition series: its high sublimation and ionisation enthalpies are not offset by its hydration enthalpy, giving a positive E∘(Cu2+/Cu)=+0.34 V, so it does not liberate H2 from acids.
Standard reduction potentials E∘(M2+/M):
- Ti: −1.63 V
- V: −1.18 V
- Cr: −0.90 V
- Fe: −0.44 V
- Cu: +0.34 V (the only positive value)
Hence copper is the answer.
✓Final answerThe correct option is (D).
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.The standard electrode potentials of some electrodes are given below: Fe3+/Fe2+ 0.77V; Br2/Br− 1.09V; I2/I− 0.54V; Zn2+/Zn(s) −0.76V; Ag+/Ag(s) 0.80V; Fe2+/Fe(s) −0.44V; Cu2+/Cu(s) 0.34V. Predict the reaction that is not feasible: (A) Fe3+(aq) oxidises I−(aq) (B) Ag+(aq) oxidises Cu(s) (C) Ag(s) reduces Fe3+(aq) (D) Br2(aq) oxidises Fe2+(aq) (E) Zn(s) reduces Cu2+(aq)
›Reveal solutionSolution
Ag (E = 0.80 V) cannot reduce Fe3+ (E = 0.77 V) because the cell EMF is negative.
Concept and Intuition
A redox reaction proceeds spontaneously only when the standard cell potential E_cell = E_reduction(oxidant) - E_reduction(reductant's own couple) is positive. A stronger oxidant (higher reduction potential) can oxidise a species with a lower reduction potential.
Step-by-Step Solution
- (A) Fe3+ (0.77) oxidises I- (0.54): E = 0.77 - 0.54 = +0.23 V, feasible.
- (B) Ag+ (0.80) oxidises Cu (0.34): E = 0.80 - 0.34 = +0.46 V, feasible.
- (C) Ag (0.80) reduces Fe3+ (0.77): E = 0.77 - 0.80 = -0.03 V, NOT feasible.
- (D) Br2 (1.09) oxidises Fe2+ (0.77): E = 1.09 - 0.77 = +0.32 V, feasible.
- (E) Zn (-0.76) reduces Cu2+ (0.34): E = 0.34 - (-0.76) = +1.10 V, feasible.
Common Mistakes
- Reversing the sign convention and concluding the wrong reaction is spontaneous.
✓Final answerThe correct option is (C) — Ag(s) reduces Fe3+(aq).
ANSWER: C
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