When the oxidation-reduction reaction shown here is balanced, how many electrons are transferred for each atom of copper that reacts?

Ag^+ + Cu = Ag + Cu^2+

Cu is zero in the metal.

It is +2 on the right.
In going from zero to +2, Cu must have .......(gained/lost) two electrons.

To balance the oxidation-reduction reaction, you need to make sure that the number of electrons lost (oxidation) is equal to the number of electrons gained (reduction).

In the given reaction, the Ag^+ ion is being reduced to Ag, and the Cu atom is being oxidized to Cu^2+.

The Ag^+ ion gains one electron to become Ag: Ag^+ + 1e^- = Ag. This is the reduction half-reaction.

The Cu atom loses two electrons to become Cu^2+: Cu - 2e^- = Cu^2+. This is the oxidation half-reaction.

To make the number of electrons lost equal to the number of electrons gained, we need to multiply the reduction reaction by 2:
2Ag^+ + 2e^- = 2Ag

Now, the number of electrons lost by the Cu atom (2) is equal to the number of electrons gained by the Ag^+ ion (2).

Therefore, for each atom of copper that reacts, 2 electrons are transferred.

To determine the number of electrons transferred for each atom of copper that reacts in this oxidation-reduction reaction, we need to balance the equation.

First, let's assign oxidation numbers to each element in the reaction:

Ag^+ + Cu = Ag + Cu^2+
(+1) (0) (0) (+2)

The oxidation number of Ag is +1 in Ag^+ and 0 in Ag. The oxidation number of Cu is 0 in Cu and +2 in Cu^2+.

Next, we'll balance the charge on each side of the equation by adding or subtracting electrons (e^-):

Ag^+ + Cu + 2e^- = Ag + Cu^2+

On the left-hand side, there are two extra electrons (2e^-) compared to the right-hand side. Therefore, each atom of copper that reacts transfers 2 electrons.

In summary, for each atom of copper that reacts, 2 electrons are transferred in this oxidation-reduction reaction.