Examine the unbalanced electrolytic reaction.

Li+(aq) + Zn(s) → Li(s) + Zn2+(aq)

How many electrons are transferred in the balanced equation?

4
1
2
3

First place, this reaction will not occur. But if it did, then

2Li^+ + Zn ==> 2Li + Zn^2+
Is 3 a guess or did you figure out it was 3? I am interested in how you solved the problem.

To determine the number of electrons transferred in the balanced equation, we need to balance the equation first. Let's start by writing the half-reactions for the oxidation and reduction processes:

Oxidation half-reaction:
Zn → Zn2+ + 2e-

Reduction half-reaction:
Li+ + e- → Li

Now we can combine the two half-reactions and balance them:

Zn + 2 Li+ → Zn2+ + 2 Li

The coefficient in front of Zn represents the number of electrons transferred. In this case, the coefficient is 2, which means that 2 electrons are transferred in the balanced equation. Therefore, the correct answer is 2.

To determine how many electrons are transferred in the balanced equation, we need to balance the equation first. The goal is to have equal numbers of atoms on both sides of the equation.

Li+(aq) + Zn(s) → Li(s) + Zn2+(aq)

First, let's examine the oxidation states of the atoms involved. Li is in its standard state, so its oxidation state is +1. Zn is a solid, so its oxidation state is 0. After the reaction, Li is in its standard state, so its oxidation state is 0. Zn is now in its +2 oxidation state.

Next, let's identify the atoms undergoing oxidation and reduction. Li goes from an oxidation state of +1 to 0, so it is being reduced. Zn goes from an oxidation state of 0 to +2, so it is being oxidized.

To balance the equation, we need to keep the total charge the same on both sides. Since Li+ has a charge of +1 and Zn2+ has a charge of +2, we can balance the charges by representing the transferred electrons.

The balanced equation is:

2 Li+(aq) + Zn(s) → 2 Li(s) + Zn2+(aq) + 2 e-

From the balanced equation, we can see that 2 electrons are transferred in the reaction. Therefore, the correct answer is 2.