JEE Main 2023 — Redox Reactions Question with Solution

To find the sum of X, Y, and Z, we need to balance the given chemical reaction:

$${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{XH}}^{+}+6{\mathrm{Fe}}^{2+}\to {\mathrm{YCr}}^{3+}+6{\mathrm{Fe}}^{3+}+{\mathrm{ZH}}_2\mathrm{O}$$

Let's balance the reaction step by step:

1. Balance the chromium atoms:

There are 2 chromium atoms on the left side of the equation, so we need 2 chromium atoms on the right side. Thus, Y = 2.

$${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{XH}}^{+}+6{\mathrm{Fe}}^{2+}\to 2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+{\mathrm{ZH}}_2\mathrm{O}$$

2. Balance the oxygen atoms:

There are 7 oxygen atoms on the left side of the equation, so we need 7 oxygen atoms on the right side. Since each water molecule (H₂O) contains one oxygen atom, Z = 7.

$${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{XH}}^{+}+6{\mathrm{Fe}}^{2+}\to 2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$$

3. Balance the hydrogen atoms and charges:

There are 14 hydrogen atoms on the right side of the equation, so we need 14 hydrogen atoms on the left side. Thus, X = 14.

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

Now the chemical reaction is balanced:

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

The sum of X, Y, and Z is:

X + Y + Z = 14 + 2 + 7 = 23

So, the sum of X, Y, and Z is 23.