Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following

Exercises · 7.26

Q. Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a) Fe3+(aq) and I–(aq)

(b) Ag+(aq) and Cu(s)

(d) Ag(s) and Fe3+(aq)

(e) Br2(aq) and Fe2+(aq).

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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 feasibility of a redox reaction is determined by the sign of the cell potential $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ . A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous. We apply this to each pair using standard electrode potentials from Table 8.1.

The key idea is simple: for any proposed redox reaction, one species gets reduced (gains electrons) and the other gets oxidised (loses electrons). The standard electrode potential tells us how strongly a species wants to be reduced. The half-reaction with the higher (more positive) reduction potential will actually undergo reduction — it acts as the cathode. The other half-reaction, with the lower reduction potential, will be forced to run in reverse (oxidation) — it acts as the anode.

The cell potential is then:

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

If $E^\circ_{\text{cell}} > 0$ , the reaction is feasible (spontaneous under standard conditions). If $E^\circ_{\text{cell}} < 0$ , it is not.

Let’s take the standard reduction potentials we need — they are in Table 7.1 of this chapter (the exercise says "Table 8.1": a leftover of the book's own pre-rationalisation chapter numbering):

$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ : $E^\circ = +0.77\ \text{V}$
$\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$ : $E^\circ = +0.54\ \text{V}$
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$ : $E^\circ = +0.80\ \text{V}$
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ : $E^\circ = +0.34\ \text{V}$
$\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$ : $E^\circ = +1.09\ \text{V}$

Now, examine each case.

Fe³⁺(aq) and I⁻(aq) The possible reaction: Fe³⁺ gets reduced to Fe²⁺, and I⁻ gets oxidised to I₂.
- Reduction (cathode): $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ , $E^\circ = +0.77\ \text{V}$
- Oxidation (anode): $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ , $E^\circ = -0.54\ \text{V}$ (reverse of the given reduction)

E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23\ \text{V} > 0

So the reaction is feasible. Fe³⁺ can oxidise I⁻ to I₂.

Ag⁺(aq) and Cu(s) Possible reaction: Ag⁺ gets reduced to Ag, Cu gets oxidised to Cu²⁺.
- Reduction: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ , $E^\circ = +0.80\ \text{V}$
- Oxidation: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$ , $E^\circ = -0.34\ \text{V}$

E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46\ \text{V} > 0

Feasible. Silver ions will oxidise copper metal.

Fe³⁺(aq) and Cu(s) Fe³⁺ can be reduced to Fe²⁺, Cu oxidised to Cu²⁺.
- Reduction: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ , $E^\circ = +0.77\ \text{V}$
- Oxidation: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$ , $E^\circ = -0.34\ \text{V}$

E^\circ_{\text{cell}} = 0.77 - 0.34 = +0.43\ \text{V} > 0

Feasible. Fe³⁺ can oxidise copper metal.

Ag(s) and Fe³⁺(aq) Here, Ag could be oxidised to Ag⁺, and Fe³⁺ reduced to Fe²⁺.
- Reduction: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ , $E^\circ = +0.77\ \text{V}$ …

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Q.Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

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