Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver el

Exercises · 7.27

Q. Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes

(ii) An aqueous solution AgNO3 with platinum electrodes

(iii) A dilute solution of H2SO4 with platinum electrodes

(iv) An aqueous solution of CuCl2 with platinum electrodes.

Nagaland NbseTextbookSubjective· 3mImportance★★★★★

47% · 37/78 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

Electrolysis predictions depend on the electrode material (active vs. inert) and the relative discharge potentials of ions present. For AgNO₃ with Ag electrodes, Ag deposits at cathode and Ag dissolves at anode. With Pt electrodes, Ag deposits at cathode but O₂ evolves at anode. For dilute H₂SO₄ with Pt, H₂ and O₂ evolve (water electrolysis). For CuCl₂ with Pt, Cu deposits and Cl₂ evolves.

The Core Idea: Why Electrolysis Predictions Work

Electrolysis is about forcing a non-spontaneous redox reaction using electrical energy. The key question is always: which ion gets reduced at the cathode, and which gets oxidised at the anode?

The answer depends on two things:

Standard reduction potentials — the more positive (or less negative) the potential, the easier it is to reduce that species.
Overpotential — some gases (especially O₂ and H₂) require extra voltage to form, which can shift the order.
Electrode material — if the anode is made of a metal like Ag or Cu, it can itself get oxidised instead of the anions in solution.

Let's apply this logic to each case.

(i) Aqueous AgNO₃ with silver electrodes

Cathode: Possible reductions:

$\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$ , $E^\circ = +0.80\ \text{V}$
$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ , $E^\circ = -0.83\ \text{V}$ (at pH 7)

Ag⁺ has a much higher reduction potential, so Ag metal deposits on the cathode.

Anode: Possible oxidations:

$\text{Ag}(s) \rightarrow \text{Ag}^+ + e^-$ , $E^\circ = -0.80\ \text{V}$ (reverse of reduction)
$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ , $E^\circ = -1.23\ \text{V}$
$\text{NO}_3^-$ is very hard to oxidise (nitrate is stable)

The silver electrode itself can oxidise more easily than water. So the anode dissolves: Ag atoms lose electrons and go into solution as Ag⁺.

Tip

With an active anode (same metal as the cation in solution), the anode dissolves and the cathode deposits the same metal. This is the principle behind electrorefining of silver.

Products: Cathode — Ag(s); Anode — Ag⁺ goes into solution (electrode dissolves).

(ii) Aqueous AgNO₃ with platinum electrodes

Cathode: Same as above — Ag⁺ reduction is favoured. Ag deposits.

Anode: Now the electrode is inert (Pt doesn't oxidise easily). So we compare:

$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ , $E^\circ = -1.23\ \text{V}$
$\text{NO}_3^-$ oxidation: nitrate is extremely stable in water; its oxidation potential is much more negative.

Water oxidation to O₂ is the only feasible reaction. So oxygen gas evolves at the anode.

Watch out

A common mistake is to think NO₃⁻ gets oxidised. In aqueous solution, nitrate ions are almost never discharged at the anode because water oxidises more easily. The same applies to sulphate, phosphate, etc.

Products: Cathode — Ag(s); Anode — O₂(g) + H⁺ (solution becomes acidic).

(iii) Dilute H₂SO₄ with platinum electrodes

This is essentially electrolysis of water with an inert electrolyte to make it conductive.

Cathode: Possible reductions:

$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ , $E^\circ = 0.00\ \text{V}$ (but in dilute solution, [H⁺] is low, so effective potential is slightly negative)
$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ , $E^\circ = -0.83\ \text{V}$

H⁺ reduction is much easier. Hydrogen gas evolves.

Anode: Possible oxidations:

$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ , $E^\circ = -1.23\ \text{V}$
$\text{SO}_4^{2-}$ oxidation: sulphate is very stable; its discharge potential is much more negative than water's.

So oxygen gas evolves.

\text{Cathode: } 2\text{H}^+ + 2e^- \rightarrow \text{H}_2

Start your 14-day free trial to unlock the full solution →

←Previous question · exercises7.26 — Using the standard electrode potentials given in the Table 8.1, predic…→Next question · exercises7.28 — Arrange the following metals in the order in which they displace each …

Q.Predict the products of electrolysis in each of the following:

The Core Idea: Why Electrolysis Predictions Work

(i) Aqueous AgNO₃ with silver electrodes

(ii) Aqueous AgNO₃ with platinum electrodes

(iii) Dilute H₂SO₄ with platinum electrodes

🎓Unlock everything free for 14 days

Unlock everything free for 14 days