JEE Main 2024 — Redox Reactions Question with Solution

• Balancing the Half-Reaction

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


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

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

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

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

1. Charge : Add 6e⁻ to the left:

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

• Identifying A :

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

• Final Balanced Equation :

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

• X = 14


• Y = 6


• Z = 7


• A = Cr³⁺