JEE MAIN - Chemistry (2024 - 1st February Morning Shift - No. 17)
In acidic medium, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ shows oxidising action as represented in the half reaction:
$$ \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O} $$
$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and $\mathrm{A}$ are respectively are :
$$ \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O} $$
$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and $\mathrm{A}$ are respectively are :
$14,7,6$ and $\mathrm{Cr}^{3+}$
$14,6,7$ and $\mathrm{Cr}^{3+}$
$8,4,6$ and $\mathrm{Cr}_2 \mathrm{O}_3$
$8,6,4$ and $\mathrm{Cr}_2 \mathrm{O}_3$
Explanation
- Balancing the Half-Reaction
- Chromium (Cr) : Already balanced with 2 Cr on each side.
- Oxygen (O): Add 7 H₂O to the right:
$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + \mathrm{XH}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$$
- Hydrogen (H) : Add 14 H⁺ to the left:
$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$$
- 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 :
$$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}$$
Answer- X = 14
- Y = 6
- Z = 7
- A = Cr³⁺
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