which of the following is reduced in the reaction:

Cu(NO3)2 + Zn --- Zn(NO3)2 + Cu

Reduction is the gain of electrons. Which gains electrons?

I'll help a little with oxidation states. On the left, Cu is +2, N is +5, O is -2, Zn is zero EACH.
On the right, Zn is +2, N is +5, O is -2, and Cu is zero

In this reaction, the copper ion (Cu²⁺) is being reduced. It's going from a positive charge of 2+ to 0. So, you could say that the copper ion is finally feeling a bit "negatively charged" about itself after getting reduced. It's like going from being the life of the party to feeling a little deflated. Tough break, copper ion!

The reaction you provided is a redox reaction, where one species is oxidized (loses electrons) while another is reduced (gains electrons). In this case, let's analyze the reaction to identify which species is reduced.

The reaction is:
Cu(NO3)2 + Zn → Zn(NO3)2 + Cu

In this reaction, copper(II) nitrate (Cu(NO3)2) and zinc (Zn) are the reactants, and zinc nitrate (Zn(NO3)2) and copper (Cu) are the products.

To identify the reduced species, we need to look at the changes in oxidation numbers. The oxidation number of an element indicates the number of electrons it has gained or lost in a compound.

In Cu(NO3)2, the oxidation number of copper (Cu) is +2, and the oxidation number of nitrogen (N) and oxygen (O) is -2.

In Zn(NO3)2, the oxidation number of zinc (Zn) is +2, and the oxidation number of nitrogen (N) and oxygen (O) is -2.

Based on the changes in oxidation numbers, we can see that the oxidation number of copper (Cu) in the reactant (Cu(NO3)2) is +2, while the oxidation number of copper (Cu) in the product (Cu) has not changed. Therefore, copper (Cu) is not reduced.

On the other hand, the oxidation number of zinc (Zn) in the reactant (Zn) is 0, while the oxidation number of zinc (Zn) in the product (Zn(NO3)2) is +2. Therefore, zinc (Zn) has been oxidized.

Since zinc (Zn) has been oxidized (increased in oxidation number), it means that copper (Cu) is reduced in the given reaction.

In order to determine which substance is reduced in the reaction, we need to identify the oxidation states of the elements before and after the reaction.

Let's start by determining the oxidation states:

Cu(NO3)2
Since NO3 is a polyatomic ion with a charge of -1, and there are two of them, it totals -2.
To find the oxidation state of Cu, we need to set up the equation:
2x + (-2) = 0 (the overall charge is zero for a neutral compound)
Solving for x, we find that the oxidation state of Cu is +2.

Zn
Zinc (Zn) is an element with no charge, so its oxidation state is 0.

Zn(NO3)2
Similar to Cu(NO3)2, NO3 is a polyatomic ion with a charge of -1, and there are two of them, so it totals -2.
To find the oxidation state of Zn, we need to set up the equation:
x + (-2) = 0 (the overall charge is zero for a neutral compound)
Solving for x, we find that the oxidation state of Zn is +2.

Now, let's compare the oxidation states before and after the reaction:

In Cu(NO3)2, the oxidation state of Cu is +2.
In Zn(NO3)2, the oxidation state of Zn is +2.

In the reaction, Cu goes from an oxidation state of +2 to 0, which means it gained electrons and was reduced. Zn remains in the oxidation state of +2, which means it did not change its oxidation state.

Therefore, in the given reaction, Cu is reduced.