E⊖ values of some redox couples are given below. On the basis of these values choose the correct option. E⊖ va

NCERT Exemplar · Q3

Q. E⊖ values of some redox couples are given below. On the basis of these values choose the correct option. E⊖ values: Br2/Br^- = +1.90; Ag^+/Ag(s) = +0.80; Cu^2+/Cu(s) = +0.34; I2(s)/I^- = +0.54

(i) Cu will reduce Br^-

(ii) Cu will reduce Ag

(iii) Cu will reduce I^-

(iv) Cu will reduce Br2

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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 $E^\ominus$ .
A common zero. Potentials can only be measured as differences, so one electrode is defined as the reference: the standard hydrogen electrode (SHE), $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ , is fixed at exactly 0.00 V. Every $E^\ominus$ is the voltage of a couple measured against it.

By convention the values are tabulated for the reduction direction:

\text{Oxidised form} + ne^- \rightarrow \text{Reduced form} \qquad E^\ominus \text{ (in volts, at 298 K)}

Reading the Table

The standard-potential table (Table 7.1 in the Class 11 chapter) runs from

$\text{F}_2/\text{F}^-$ at $+2.87$ V down to $\text{Li}^+/\text{Li}$ at $-3.05$ V.

Two rules unlock it:

More positive $E^\ominus$ → stronger oxidising agent (the oxidised form is hungrier for electrons). F₂ tops the table; that is why fluorine oxidises almost everything.
More negative $E^\ominus$ → 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 $E^\ominus$ 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:

E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}

Worked feel: can Fe³⁺ oxidise iodide? $E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V is above $E^\ominus(\text{I}_2/\text{I}^-) = +0.54$ V, so

$E^\ominus_{\text{cell}} = +0.23$ 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 $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ with BOTH values as tabulated (reduction) potentials — the subtraction already handles the reversal.
Multiplying $E^\ominus$ by stoichiometric coefficients. Potentials are intensive: doubling a half reaction does not double its $E^\ominus$ .
Reading "negative" as "impossible". A negative $E^\ominus$ 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 $E^\ominus$ 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 $E^\ominus$ to non-standard concentrations, and cell EMF

connects to thermodynamics via $\Delta G = -nFE$ .

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):

\text{M}_1 \rightarrow \text{M}_1^{n+} + n e^-

M₂ acts as the cathode — it reduces something (e.g., oxygen or H⁺):

\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^- \quad \text{(in neutral/alkaline)}

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad \text{(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 $E_{\text{corr},1}$ ; for M₂, it's $E_{\text{corr},2}$ .

When connected, the system cannot stay at two different potentials. The entire metal couple must reach a single potential — the mixed potential $E_{\text{mix}}$ .

$E_{\text{mix}}$ lies between $E_{\text{corr},1}$ and $E_{\text{corr},2}$ .
At $E_{\text{mix}}$ , the total anodic current from M₁ equals the total cathodic current from M₂ (charge conservation):

I_{\text{anode}} = I_{\text{cathode}}

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 $i_a$ at potential $E$ is:

i_a = i_{0,1} \exp\left( \frac{\alpha_a F (E - E_{0,1})}{RT} \right)

For the cathode (M₂), the cathodic current density $i_c$ is:

i_c = i_{0,2} \exp\left( -\frac{\alpha_c F (E - E_{0,2})}{RT} \right)

Where:

$i_{0,1}, i_{0,2}$ = exchange current densities
$\alpha_a, \alpha_c$ = transfer coefficients (typically ~0.5)
$F$ = Faraday constant
$R$ = gas constant
$T$ = temperature
$E_{0,1}, E_{0,2}$ = standard reduction potentials

At the mixed potential $E_{\text{mix}}$ :

I_{\text{galvanic}} = A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

Where $A_1$ and $A_2$ 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:

\text{Corrosion rate} = \frac{I_{\text{galvanic}} \cdot M}{n \cdot F \cdot \rho}

Where:

$M$ = molar mass of the anode metal
$n$ = number of electrons transferred per atom
$\rho$ = 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 $Q = I_{\text{galvanic}} \cdot t$ corresponds to moles of metal lost:

\text{moles lost} = \frac{Q}{nF} = \frac{I_{\text{galvanic}} \cdot t}{nF}

Multiply by $M/\rho$ to get volume or thickness loss.

5. The Area Effect: Why It Matters

From the mixed potential equation:

A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

If the cathode area $A_2$ is large relative to the anode area $A_1$ , then $i_a(E_{\text{mix}})$ 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 $i_a = I_{\text{galvanic}} / A_1$ . For a fixed $I_{\text{galvanic}}$ , smaller $A_1$ gives higher $i_a$ , which accelerates corrosion.

6. The Driving Force: Potential Difference

The driving force for galvanic corrosion is the difference in open-circuit potentials:

\Delta E = E_{\text{corr},2} - E_{\text{corr},1}

A larger $\Delta E$ generally leads to a larger $I_{\text{galvanic}}$ , but the exact relationship depends on the polarization behavior (Tafel slopes) of both electrodes.

Why this holds: The mixed potential $E_{\text{mix}}$ 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	$I_{\text{anode}} = I_{\text{cathode}}$	Charge conservation in a closed circuit
Galvanic current	$A_1 i_a(E_{\text{mix}}) = A_2 i_c(E_{\text{mix}})$	Butler-Volmer kinetics + area balance
Corrosion rate	$\frac{I_{\text{galvanic}} M}{n F \rho}$	Faraday's law of electrolysis
Area effect	Small anode → high $i_a$	Current density inversely proportional to area
Driving force	$\Delta E = E_{\text{corr},2} - E_{\text{corr},1}$	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.

A species can reduce another if it has a lower (more negative) standard electrode potential. Copper metal ( $E^\ominus = +0.34\,\text{V}$ ) can reduce only those species with higher potentials; among the options, only $\text{Br}_2$ ( $E^\ominus = +1.90\,\text{V}$ ) qualifies. The correct option is (iv).

The question hinges on understanding spontaneous redox reactions through standard electrode potentials. A redox reaction is spontaneous when the overall cell potential $E^\ominus_\text{cell}$ is positive. This happens when electrons flow from a species that loses them easily (lower $E^\ominus$ , better reducing agent) to one that accepts them readily (higher $E^\ominus$ , better oxidising agent).

When we say "Cu will reduce X," we mean copper metal acts as the reducing agent—it gets oxidised to $\text{Cu}^{2+}$ while X gets reduced. For this to be spontaneous:

E^\ominus_\text{cell} = E^\ominus_\text{cathode} - E^\ominus_\text{anode} > 0

Here the cathode is where X gets reduced, and the anode is where Cu gets oxidised. Rearranging, we need:

E^\ominus(\text{X}/\text{reduced form of X}) > E^\ominus(\text{Cu}^{2+}/\text{Cu})

In other words, the species being reduced must have a higher standard potential than copper.

Let's examine each option systematically.

1. Option (i): Cu will reduce $\text{Br}^-$

The proposed half-reactions would be:

Oxidation: $\text{Cu}(s) \to \text{Cu}^{2+} + 2e^-$ with $E^\ominus = +0.34\,\text{V}$
Reduction: $\text{Br}^- \to ?$

But $\text{Br}^-$ is already in its reduced form. To "reduce" it further is impossible—it would need to accept more electrons, but bromide ion is stable and cannot be reduced under standard conditions. The reverse reaction, oxidising $\text{Br}^-$ to $\text{Br}_2$ , would require an oxidising agent stronger than copper. This option makes no chemical sense.

2. Option (ii): Cu will reduce Ag

Again, $\text{Ag}(s)$ is already the reduced form of silver. Copper cannot reduce metallic silver further. What can happen is the reverse: $\text{Ag}^+$ can oxidise copper, because $E^\ominus(\text{Ag}^+/\text{Ag}) = +0.80\,\text{V} > +0.34\,\text{V}$ . But that's not what the option states.

3. Option (iii): Cu will reduce $\text{I}^-$

Like bromide, iodide ion is already reduced. Copper cannot reduce $\text{I}^-$ further. (Copper could be oxidised by $\text{I}_2$ , but that's the opposite process.)

4. Option (iv): Cu will reduce $\text{Br}_2$

Now we have a genuine redox pair. The half-reactions are:

Oxidation: $\text{Cu}(s) \to \text{Cu}^{2+} + 2e^-$ with $E^\ominus_\text{anode} = +0.34\,\text{V}$
Reduction: $\text{Br}_2 + 2e^- \to 2\text{Br}^-$ with $E^\ominus_\text{cathode} = +1.90\,\text{V}$

The cell potential is:

E^\ominus_\text{cell} = 1.90 - 0.34 = +1.56\,\text{V}

Since $E^\ominus_\text{cell} > 0$ , the reaction is spontaneous. Copper metal will indeed reduce bromine.

Tip

A quick mnemonic: Higher potential wins the electrons. The species with the higher $E^\ominus$ acts as the oxidising agent (gets reduced), while the one with lower $E^\ominus$ acts as the reducing agent (gets oxidised).

Watch out

Don't confuse the reduced form with the ability to be reduced further. $\text{Br}^-$ , $\text{Ag}(s)$ , and $\text{I}^-$ are already reduced; they cannot accept more electrons. Only their oxidised counterparts ( $\text{Br}_2$ , $\text{Ag}^+$ , $\text{I}_2$ ) can be reduced.

✓Final answer

\text{Br}_2

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Q.E⊖ values of some redox couples are given below. On the basis of these values choose the correct option. E⊖ values: Br2/Br^- = +1.90; Ag^+/Ag(s) = +0.80; Cu^2+/Cu(s) = +0.34; I2(s)/I^- = +0.54

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Q.E⊖ values of some redox couples are given below. On the basis of these values choose the correct option.
E⊖ values: Br2/Br^- = +1.90; Ag^+/Ag(s) = +0.80; Cu^2+/Cu(s) = +0.34; I2(s)/I^- = +0.54

Unlock everything free for 14 days