QJEEM24Che331: JEE Main 2024 Chemistry - Class 11 JEEM24Che331 Solution: In acidic medium, shows oxidising action as represented in the…

• Balancing the Half-Reaction

1. Chromium (Cr) : Already balanced with 2 Cr on each side.


2. Oxygen (O): Add 7 H₂O to the right:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + \mathrm{XH}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

1. Hydrogen (H) : Add 14 H⁺ to the left:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

1. Charge : Add 6e⁻ to the left:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

• Identifying A :

The reduction of Cr₂O₇²⁻ in acidic medium forms Cr³⁺ ions.

• Final Balanced Equation :

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}$

• X = 14


• Y = 6


• Z = 7


• A = Cr³⁺