Question

X and Y are the number of electrons involved, respectively during the oxidation of I$^-$ to I$_2$ and S$^{2-}$ to S by acidified K$_2$Cr$_2$O$_7$. The value of X + Y is ___.

Hint:

In redox reactions, always balance electrons first to determine the number of electrons transferred accurately.

Answer 1

Correct Answer: 4

To solve the problem, we must determine the number of electrons transferred in the oxidation reactions of iodide (I^-) to iodine (I₂) and sulfide (S^2-) to sulfur when reacted with acidified potassium dichromate (K₂Cr₂O₇).

Oxidation of I^- to I₂:
In this reaction, each iodide ion loses one electron:
I^- → I₂ + 2e^-
For the formation of one molecule of I₂, two electrons are lost.
Thus, X = 2.

Oxidation of S^2- to S:
In this reaction, each sulfide ion loses two electrons:
S^2- → S + 2e^-
As one atom of sulfur is formed, two electrons are lost.
Thus, Y = 2.

Adding the electrons involved in both reactions:
X + Y = 2 + 2 = 4.

The value of X + Y is 4, which falls within the range [4, 4].