Use oxidation states to identify the element that is being oxidized in the following redox reaction:

Cu(s)+2
H
2
S
O
4
(aq) → CuS
O
4
(aq)+S
O
2
(g)+2
H
2
O(l)

You would make a lot more sense if you wrote it this way.

Cu + 2H2SO4 ==>CuSO4 + SO2 + 2H2O

Remember the rules.
Oxidation is the loss of electrons.
Cu is 0 on the left and +2 on the right.
S is +6 on the left and +4 on the right (with SO2)
H is +1 on left and right(doesn't change)
O is -2 on left and right(doesn't change)

Well, let's break this one down. The oxidation state of copper (Cu) in its elemental state (Cu(s)) is 0. In the product, CuSO4, the oxidation state of copper is +2. So, that means copper has been oxidized from an oxidation state of 0 to +2. Hence, the element being oxidized in this redox reaction is copper.

Now, I know what you're thinking - "Why did the copper go to therapy? Because it was tired of feeling Cu-pid!"

To identify the element being oxidized in a redox reaction, we can assign oxidation states to each element and observe the changes in these states.

In the given redox reaction:
Cu(s) + 2 H2SO4(aq) → CuSO4(aq) + SO2(g) + 2 H2O(l)

We need to compare the oxidation states of the elements before and after the reaction:

For Cu (copper):
The oxidation state of Cu in its elemental state is 0 since it is not bonded to any other element.
The oxidation state of Cu in CuSO4 is +2 since it is bonded to four oxygen atoms, each having an oxidation state of -2, which sums up to the oxidation state of +2 for Cu.

For S (sulfur):
The oxidation state of S in H2SO4 is +6 since each O atom has an oxidation state of -2, and each H atom has an oxidation state of +1.
The oxidation state of S in CuSO4 is +6 as well since the oxidation state of Cu is +2.

For O (oxygen):
The oxidation state of O in H2SO4 is -2 since it is combined with two H atoms.
The oxidation state of O in CuSO4 is -2 since it is bonded to Cu, whose oxidation state is +2.

Therefore, the element being oxidized in the reaction is sulfur (S). The oxidation state of sulfur is reduced from +6 in H2SO4 to +4 in CuSO4.

To identify the element that is being oxidized in a redox reaction, we need to determine the changes in oxidation states of the elements involved.

In this redox reaction:
Cu(s) + 2 H2SO4(aq) → CuSO4(aq) + SO2(g) + 2 H2O(l)

First, let's assign the oxidation states to the elements based on the common oxidation state rules and considering the charges of certain ions:
- The oxidation state of Cu(s) is 0 since it is an uncombined element.
- The oxidation state of H in H2SO4 is +1.
- The oxidation state of O in H2SO4 is -2.
- The oxidation state of S in H2SO4 is +6.

Now, we can track the changes in oxidation states:
- In CuSO4(aq), the oxidation state of Cu increases from 0 to +2.
- In SO2(g), the oxidation state of S decreases from +6 to +4.
- In H2O(l), the oxidation state of H remains +1.
- In H2SO4(aq), the oxidation states of H and O remain the same.

Based on the changes in oxidation states, we can conclude that the element being oxidized in this redox reaction is copper (Cu). The oxidation state of Cu increased from 0 to +2.

Therefore, in the given redox reaction, copper (Cu) is being oxidized.