assuming the following reaction proceeds in the forward direction, what is oxidized and what is reduced?

Cu2+ (aq) + Ca(s) --> Cu (s) + Ca2+ (aq)

Oxidation is the loss of electrons.

Reduction is the gain of electrons.
Cu goes from +2 on the left to 0 on the right.
Ca goes from 0 on the left to +2 on the right.

In the given reaction, Cu2+ (aq) + Ca(s) → Cu (s) + Ca2+ (aq), copper (Cu2+) is being reduced, and calcium (Ca) is being oxidized.

To determine oxidation and reduction, we look at the change in oxidation numbers of the elements involved in the reaction.

Initially, copper (Cu) has an oxidation state of +2 (Cu2+). In the product, copper (Cu) has an oxidation state of 0 (Cu).

Initially, calcium (Ca) has an oxidation state of 0 (Ca). In the product, calcium (Ca) has an oxidation state of +2 (Ca2+).

The oxidation state of copper changes from +2 to 0, indicating a gain of electrons, and hence, it is being reduced.

On the other hand, the oxidation state of calcium changes from 0 to +2, indicating a loss of electrons, and thus, it is being oxidized.

Therefore, copper (Cu2+) is reduced, and calcium (Ca) is oxidized in this reaction.

To determine what is oxidized and what is reduced in a chemical reaction, you need to compare the oxidation states of the elements involved in the reaction.

In this reaction, copper (Cu) is reduced from an oxidation state of +2 to 0, while calcium (Ca) is oxidized from an oxidation state of 0 to +2.

Here's how to determine this:

1. Identify the oxidation states of all elements in the reactants and products. In this case, copper (Cu) has an oxidation state of +2 in Cu2+ (aq) and 0 in Cu (s), while calcium (Ca) has an oxidation state of 0 in Ca(s) and +2 in Ca2+ (aq).

2. Compare the oxidation states before and after the reaction. Copper undergoes a decrease in oxidation state from +2 to 0, which means it gains electrons and is reduced (since it has a more negative charge afterward).

3. Calcium, on the other hand, undergoes an increase in oxidation state from 0 to +2, which means it loses electrons and is oxidized (since it has a more positive charge afterward).

Therefore, in the given reaction, copper (Cu2+) is reduced, and calcium (Ca) is oxidized.