E^°cell for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or m

NCERT Exemplar · Q44

E^\circ_{cell}

(a)

H^+(aq) + e^- \rightarrow \frac{1}{2}H_2(g)

;

E^\circ_{cell} = 0.00\ V

(b)

2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-

;

E^\circ_{cell} = 1.23\ V

(c)

2SO_4^{2-}(aq) \rightarrow S_2O_8^{2-}(aq) + 2e^-

;

E^\circ_{cell} = 1.96\ V

(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.

(iii) In dilute sulphuric acid solution, water will be oxidised at anode.

(iv) In dilute sulphuric acid solution,

SO_4^{2-}

ion will be oxidised to tetrathionate ion at anode.

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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 key idea is that in electrolysis, the species with the lower reduction potential gets reduced at the cathode, and the species with the lower oxidation potential (i.e., the one that is hardest to oxidise) gets oxidised at the anode. For dilute $H_2SO_4$ , water oxidises at the anode ( $E^\circ = 1.23\ V$ ) before $SO_4^{2-}$ ( $E^\circ = 1.96\ V$ ), and $H^+$ reduces at the cathode. For concentrated $H_2SO_4$ , the effective concentration changes the competition — water oxidation becomes harder, so $SO_4^{2-}$ oxidation can occur. The correct options are (i) and (iii).

This is a classic electrolysis problem from electrochemistry. The given half-cell reactions are standard reduction potentials — but note that reaction (b) and (c) are written as oxidations in the problem statement. That’s a deliberate twist. Let’s first convert everything to a consistent language.

The core principle: In an electrolytic cell, the cathode is where reduction happens (gain of electrons), and the anode is where oxidation happens (loss of electrons). The cell is driven by an external voltage, so the reaction that occurs is not spontaneous — we force it. Which reaction actually takes place at each electrode depends on the competition among all species present.

For reduction at the cathode: the species with the higher (more positive) reduction potential gets reduced first — because it is easier to reduce.

For oxidation at the anode: the species with the lower (less positive) reduction potential (i.e., the one that is easiest to oxidise) gets oxidised first. Equivalently, look at the oxidation potentials (reverse of reduction potentials): the species with the higher oxidation potential gets oxidised first.

Let’s rewrite the given data as standard reduction potentials (all in one direction):

$2H^+ + 2e^- \rightarrow H_2$ ; $E^\circ = 0.00\ V$
$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ; $E^\circ = +1.23\ V$ (reverse of given)
$S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$ ; $E^\circ = +1.96\ V$ (reverse of given)

Now, in an aqueous solution of sulphuric acid ( $H_2SO_4$ ), the species present are: $H^+$ , $SO_4^{2-}$ , $H_2O$ , and also $OH^-$ (but in acidic solution, $OH^-$ concentration is negligible). At the cathode, possible reductions are:

$2H^+ + 2e^- \rightarrow H_2$ ( $E^\circ = 0.00\ V$ )
$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ( $E^\circ = -0.83\ V$ in neutral, but in acid it’s even less favourable)

Clearly, $H^+$ reduction has a much higher reduction potential (0.00 V) than water reduction (−0.83 V). So hydrogen ions are reduced at the cathode in both dilute and concentrated acid. That makes option (i) correct.

At the anode, possible oxidations are:

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ (reverse of reaction 2); $E^\circ_{ox} = -1.23\ V$ (since reduction potential is +1.23 V, oxidation potential is −1.23 V)
$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$ (reverse of reaction 3); $E^\circ_{ox} = -1.96\ V$

The more positive the oxidation potential, the easier the oxidation. Here, −1.23 V is greater than −1.96 V, so water oxidation is easier than sulphate oxidation. Therefore, in dilute sulphuric acid, water gets oxidised at the anode, producing oxygen gas. That makes option (iii) correct and option (iv) incorrect. …

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Method: Electrode Potential Comparison for Electrolysis

This method uses standard reduction potentials to predict which species gets oxidised (at anode) and which gets reduced (at cathode) during electrolysis.

Key rule:

Cathode (reduction): The species with the higher (more positive) reduction potential gets reduced.
Anode (oxidation): The species with the lower (less positive) reduction potential gets oxidised (reverse the sign for oxidation potential).

Step-by-step solution

Step 1: Write all half-reactions as reductions with their $E^\circ$ values

Reduction half-reaction	$E^\circ$ (V)
$2H^+ + 2e^- \rightarrow H_2$	$0.00$
$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$	$+1.23$
$S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$	$+1.96$

Step 2: Identify possible reactions at each electrode in dilute $H_2SO_4$

At cathode (reduction):

$H^+$ reduction: $E^\circ = 0.00\ V$
$H_2O$ reduction: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ; $E^\circ = -0.83\ V$ (not given, but known)
Higher $E^\circ$ wins: $H^+$ reduction ( $0.00\ V$ ) > water reduction ( $-0.83\ V$ )
✓ Hydrogen is reduced at cathode → Option (i) is correct.

At anode (oxidation):

Reverse the given reduction potentials to get oxidation potentials:

$H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ; oxidation potential = $-1.23\ V$
$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$ ; oxidation potential = $-1.96\ V$
Higher (less negative) oxidation potential wins: $-1.23\ V > -1.96\ V$
✓ Water gets oxidised at anode → Option (iii) is correct. …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing Reduction Potential with Oxidation Potential

The error: Students treat the given $E^\circ_{cell}$ values as if they are all reduction potentials. Reaction (b) and (c) are written as oxidation half-reactions, but their $E^\circ$ values are still given as positive numbers. This leads to wrong comparisons.

How to avoid:

Always check the direction of the arrow. If electrons are on the right side, it's an oxidation half-reaction.
For oxidation, the actual potential is the negative of the given value when comparing with reduction potentials.
Correct approach: Convert all to reduction potentials:
- (a) $H^+ + e^- \rightarrow \frac12 H_2$ ; $E^\circ_{red} = 0.00\ V$
- (b) $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ; $E^\circ_{red} = +1.23\ V$
- (c) $S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$ ; $E^\circ_{red} = +1.96\ V$

Mistake 2: Forgetting the Effect of Concentration on Electrode Potential

The error: Students assume the same reaction occurs at both dilute and concentrated $H_2SO_4$ , ignoring that concentration changes the actual potential via the Nernst equation.

How to avoid:

Remember: Higher concentration of $H^+$ makes $H^+$ reduction easier (more positive potential).
In dilute $H_2SO_4$ , $[H^+]$ is low → $H^+$ reduction potential is less than 0.00 V.
In concentrated $H_2SO_4$ , $[H^+]$ is high → $H^+$ reduction potential is greater than 0.00 V.
Key insight: At the anode, the species with the lowest oxidation potential (most negative or least positive) gets oxidised first.

Mistake 3: Misidentifying Which Species Gets Oxidised at the Anode

The error: Students think $SO_4^{2-}$ oxidation (option D) happens in dilute acid because its $E^\circ$ is high, without comparing with water oxidation.

How to avoid:

At the anode, oxidation occurs. Compare oxidation potentials (reverse of reduction potentials):
- Water oxidation: $E^\circ_{ox} = -1.23\ V$
- $SO_4^{2-}$ oxidation: $E^\circ_{ox} = -1.96\ V$
More positive oxidation potential means easier oxidation.
$-1.23 > -1.96$ , so water oxidises more easily than $SO_4^{2-}$ .
Correct conclusion: In dilute acid, water oxidation occurs at anode → option (iii) is correct, not (iv).

Mistake 4: Assuming Option (ii) Must Be Correct Just Because SO4^2- Oxidation Becomes Possible in Concentrated Acid

The error: Students reason that since $SO_4^{2-}$ oxidation becomes competitive in concentrated acid, option (ii) — which claims water is oxidised in concentrated acid — must be the correct description of what happens there.

Why it's wrong:

In concentrated $H_2SO_4$ , the activity of free water is drastically reduced while $[SO_4^{2-}]$ is very high. This shifts the competition away from water and toward sulphate — it is $SO_4^{2-}$ that gets oxidised to $S_2O_8^{2-}$ (peroxodisulphate) at the anode in concentrated acid, not water. Option (ii) states the opposite of this (that water is oxidised in concentrated acid), so option (ii) is actually incorrect, not correct.

How to avoid:

For each scenario (dilute vs concentrated), determine both half-reactions and check exactly what species each option names:
- Dilute $H_2SO_4$ :
  - Cathode: $H^+$ reduction (since $E^\circ_{red} \approx 0$ and no better option) …

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