Consider the following reaction equation: SO42-(aq) + 2H+(aq) + ye- \(\to\) SO42-(aq) + H2O(1). The value of y in the equation is

The given reaction equation represents the reduction of sulfate ion (SO₄^2-) to hydrogen sulfide (H₂S) using two hydrogen ions (H⁺) and some number of electrons (e^-). In order to balance the charges on both sides of the equation, the number of electrons must be equal to the positive charges on the left side. On the left side, there are 2 hydrogen ions (2H⁺) with a charge of +2 and one sulfate ion (SO₄^2-) with a charge of -2. Thus, the total charge on the left side is 0. On the right side, there is one water molecule (H₂O) with a charge of 0 and one sulfate ion (SO₄^2-) with a charge of -2. Thus, the total charge on the right side is also -2. Since the charges on both sides must be equal, the number of electrons (y) involved in the reaction must be equal to the difference between the charges on both sides, which is 2. Therefore, the correct answer is y = 2.