$E^\circ_{cell}$ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)
(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)

Question

$E^\circ_{cell}$ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)
(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.

Accepted Answer

> [!TLDR]
> 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)**.

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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):

1. $2H^+ + 2e^- \rightarrow H_2$; $E^\circ = 0.00\ V$  
2. $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$; $E^\circ = +1.23\ V$ (reverse of gi

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

Q. $E^\circ_{cell}$ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)

Concept understanding — Standard Electrode Potentials

Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"

The Core Idea

Reading the Table

Predicting Whether a Reaction Goes

Why this formula?

Galvanic Corrosion: Why the Key Formulas Hold

1. The Core Idea: Two Electrodes, One Circuit

2. The Mixed Potential: Why It Exists

3. Deriving the Galvanic Current

4. The Corrosion Rate Formula

5. The Area Effect: Why It Matters

Method: Electrode Potential Comparison for Electrolysis

Step-by-step solution

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

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

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Reduction Potential with Oxidation Potential

Mistake 2: Forgetting the Effect of Concentration on Electrode Potential

Mistake 3: Misidentifying Which Species Gets Oxidised at the Anode

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

Concept and Intuition

Step-by-Step Reasoning

1. Reaction I: Phenol + SOCl₂ → Chlorobenzene

2. Reaction II: Chlorobenzene + H₂O at 300 K → Phenol

Q.Ecell∘E^\circ_{cell}Ecell∘​ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)

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Q. $E^\circ_{cell}$ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)

Unlock everything free for 14 days