Determine the total number of electrons transferred in the following balanced reaction.

4 MnO4– (aq) + 12 H3O+ (aq) → 4 Mn2+ (aq) + 5 O2 (g) + 18 H2O (g)

Mn in MnO4^- goes from +7 on the left to +2 on the right which is a change of 5 electrons. You have 4 moles so 4 x 5 20 electron change.

Well, let's take a closer look at this chemical reaction. It might seem a bit intimidating, but don't worry – I'm here to help!

In this reaction, you can see that there are four manganese (Mn) atoms on the left side of the equation and four manganese (Mn) atoms on the right side. This means the number of manganese (Mn) atoms remains the same, and no electrons are transferred involving manganese.

Now, let's turn our attention to oxygen (O). On the left side, there are 16 oxygen (O) atoms in the form of four permanganate ions (MnO4–). However, on the right side, there are only five oxygen (O) atoms in the form of O2 gas. This means 16 - 5 = 11 oxygen (O) atoms are lost in the reaction.

Each oxygen (O) atom typically has a charge of -2, so to balance the loss of 11 oxygen (O) atoms, an equal number of electrons should be gained. Therefore, 11 * 2 = 22 electrons are transferred in relation to oxygen (O).

So, in total, we have 22 electrons transferred in this balanced reaction.

To determine the total number of electrons transferred in the balanced reaction, we need to consider the changes in oxidation states for the elements involved.

In this reaction, Mn changes from a +7 oxidation state in MnO4– to a +2 oxidation state in Mn2+.

Thus, the oxidation state of Mn changes by 7 - 2 = 5.

One MnO4– ion is involved in the reaction, so we multiply the change in oxidation state by the coefficient of MnO4– in the balanced equation, which is 4:

5 × 4 = 20

Therefore, a total of 20 electrons are transferred in the balanced reaction.

To determine the total number of electrons transferred in a balanced reaction, you need to look at the changes in oxidation states of the elements involved.

In the given reaction:
4 MnO4– (aq) + 12 H3O+ (aq) → 4 Mn2+ (aq) + 5 O2 (g) + 18 H2O (g)

The oxidation state of Mn changes from +7 on the left side to +2 on the right side, which means each Mn atom gained 5 electrons (7 - 2 = 5). Since there are 4 Mn atoms involved, the total number of electrons transferred due to Mn is 4 × 5 = 20.

On the right side of the reaction, the O atoms in MnO4– have an oxidation state of -2, while the O atoms in O2 have an oxidation state of 0. So, each O atom in the MnO4– gained 2 electrons. There are a total of 16 O atoms involved (4 MnO4–), so the total number of electrons transferred due to O is 16 × 2 = 32.

Therefore, the total number of electrons transferred in the reaction is 20 + 32 = 52.