For the oxidation-reduction reaction equation given here,

2Rb+Br2 --> 2RbBr

Indicate how many electrons are transferred in the formation of one formula unit of product.

Well, let me calculate that for you.

So, we have 2 Rb atoms on the left side, and 2 RbBr molecules on the right side. Since each Rb atom loses one electron to become Rb+, and each Br2 molecule gains two electrons to become 2 Br^-, we have a total of 2 electrons transferred per formula unit.

So, in the formation of one formula unit of RbBr, we have a "2-electron circus" going on. Be prepared for a shocking show!

To determine the number of electrons transferred in the formation of one formula unit of product in the oxidation-reduction (redox) reaction equation, we first need to identify the oxidation states of the elements involved.

In the equation 2Rb + Br2 → 2RbBr:

- Rb is a Group 1 alkali metal, which has a +1 oxidation state.
- Br is a halogen with an oxidation state of -1.

In the reactant side, Rb is oxidized from an oxidation state of 0 (as an element) to +1, while Br2 is reduced from an oxidation state of 0 (as a diatomic molecule) to -1 in RbBr.

To balance the net charge on both sides of the equation, it is necessary for the reacting species to exchange the same number of electrons. In this case, since two Rb atoms are involved, it means that two electrons are transferred in the formation of one formula unit of RbBr.

Therefore, two electrons are transferred in the formation of one formula unit of RbBr.

To determine the number of electrons transferred in the formation of one formula unit of product in the given oxidation-reduction reaction equation, you will need to examine the changes in the oxidation states of the atoms involved.

In this equation, Rb (Rubidium) has an oxidation state of +1, and Br (Bromine) has an oxidation state of 0 in the elemental form (Br2) and -1 in the compound form (RbBr).

To find the change in oxidation state for each atom, you subtract the final oxidation state from the initial oxidation state.

For Rb: +1 - 0 = +1
For Br: -1 - 0 = -1

From this, we can see that each Rb atom loses one electron while each Br atom gains one electron. Since the compound RbBr has two formula units in the balanced equation, we can conclude that two electrons are transferred in the formation of one formula unit of product (RbBr).

Rb ==> Rb^+ + e

1 mol RbBr contains 1 mol Rb metal which undergoes 1 e per atom change; therefore, 1 mole of electrons is changed for 1 mol RbBr. 1 mol electrons = 6.022E23 individual electrons.