Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hyd

Exercises · 7.15

Q. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

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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, …

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). …

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The oxidising power of halogens depends on their tendency to gain electrons (reduction potential), which decreases down the group — fluorine is the strongest oxidant. The reducing power of hydrohalic acids depends on the ease with which the H–X bond breaks to release H⁺ and X⁻, and on the ability of X⁻ to donate electrons — HI is the strongest reductant because the I⁻ ion is large, polarisable, and the H–I bond is weakest.

Why this question matters

This is a classic comparison that tests your understanding of periodic trends applied to two different but related families: the halogens (X₂) and their hydrogen compounds (HX). The key is to realise that oxidising power and reducing power are opposite sides of the same coin — but the factors that govern them are not identical.

For halogens, we look at how easily they gain an electron to become X⁻. For hydrohalic acids, we look at how easily they lose a hydrogen (as H⁺) and how readily the halide ion X⁻ can give up an electron.

Let’s break it down step by step.

1. Oxidising power of halogens: why fluorine is the best

An oxidising agent itself gets reduced — it gains electrons. For a halogen X₂, the reduction half-reaction is:

\frac{1}{2} X_2 + e^- \rightarrow X^-

The standard reduction potential $E^\circ$ measures the tendency for this to happen. A more positive $E^\circ$ means a stronger oxidant.

E^\circ (\text{V}) \text{ for } \frac{1}{2}X_2 + e^- \rightarrow X^-: \quad F_2 = +2.87,\; Cl_2 = +1.36,\; Br_2 = +1.09,\; I_2 = +0.54

Fluorine has the highest reduction potential. Why?

Small atomic size: Fluorine is the smallest halogen. Its nucleus holds incoming electrons very tightly.
Weak F–F bond: lone-pair repulsion between the two small atoms makes the F–F bond surprisingly weak, so little energy is spent breaking it.
Very high hydration enthalpy of F⁻: the tiny fluoride ion binds water exceptionally strongly. These two factors dominate the reduction potential — and they more than compensate for the fact that fluorine's electron gain enthalpy is actually slightly less negative than chlorine's.
The net result: F₂ is so eager to gain electrons that it oxidises almost everything, including water.

Watch out

A common mistake is to think that because F₂ has the strongest bond (it doesn’t — Cl₂ has a stronger bond), it should be the weakest oxidant. Actually, the bond strength trend is irregular: F–F is weaker than Cl–Cl. But the reduction potential is the direct measure of oxidising power, and it clearly decreases down the group.

So the trend is:

\text{Oxidising power: } F_2 > Cl_2 > Br_2 > I_2

Fluorine is the best oxidant.

2. Reducing power of hydrohalic acids: why HI is the best

A reducing agent itself gets oxidised — it loses electrons. For a hydrohalic acid HX, the relevant process is:

2 HX \rightarrow X_2 + 2 H^+ + 2 e^-

But more directly, we consider the halide ion X⁻ as the reducing species:

2 X^- \rightarrow X_2 + 2 e^-

The ease with which X⁻ loses an electron depends on:

Size of X⁻: Larger ions have electrons farther from the nucleus, held less tightly, so they are easier to oxidise.
Bond strength of H–X: To act as a reductant, HX must first dissociate into H⁺ and X⁻. A weaker H–X bond makes this easier.
Hydration enthalpy: A smaller hydration enthalpy (less stabilisation of X⁻ in water) makes it easier to remove an electron — but this is a secondary effect.

Let’s look at the data:

HX	H–X bond dissociation energy (kJ/mol)	Ionic radius of X⁻ (pm)	Standard oxidation potential of X⁻ (V)
HF	570	133	−2.87 (very hard to oxidise)
HCl	431	181	−1.36
HBr	366	196	−1.09

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Q.Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Why this question matters

1. Oxidising power of halogens: why fluorine is the best

2. Reducing power of hydrohalic acids: why HI is the best

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