Chemistry · d and f Block Elements

JEE Main 2025 — 23 January, Morning Shift — Question 30

$\mathrm{FeO_4^{2-}} \xrightarrow{+2.0V} \mathrm{Fe^{3+}} \xrightarrow{+0.8V} \mathrm{Fe^{2+}} \xrightarrow{-0.5V} \mathrm{Fe^{0}}$

In the above diagram, the standard electrode potentials are given in volts(over the arrow)

$\text{The value of } E^\ominus_{\mathrm{FeO_4^{2-}/Fe^{2+}}} \text{ is}$

Option A:
1.7 V
Correct
Option B:
1.2 V
Option C:
2.1 V
Option D:
1.4 V

Answer: A

Step-by-step solution

$\text{FeO}_4^{-2} \xrightarrow{E_1^\circ = 2V, n_1=3} \text{Fe}^{+3} \xrightarrow{E_2^\circ = 0.8V, n_2=1} \text{Fe}^{+2} \xrightarrow{E_3^\circ = -0.05V} \text{Fe}$ $E_4^\circ = ?$ $n_4 = 4$ $\Delta G_4^\circ = \Delta G_1^\circ + \Delta G_2^\circ$ $-n_4FE_4^\circ = -n_1FE_1^\circ - n_2FE_2^\circ$ $\Rightarrow 4E_4^\circ = 3 \times 2 + (1 \times 0.8)$ $\Rightarrow E_4^\circ = \frac{6.8}{4} \text{V}$ $\Rightarrow \boxed{E_4^\circ = 1.7 \text{V}}$

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: d and f Block Elements
Topic: Introduction and Properties of Transition Elements

The correct set of ions (aqueous solution) with same colour from the following is :

Answer: A

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