Write the overall cell reaction for a cell containing aluminum metal and Zinc(II)sulfate

so would it be:
Al (s) + ZnSO4 (aq) <---> Al3+ + ZnSO4 ??????
I tried to do it myself but I'm a bit confused...
I'm confused about the charges because isn't ZnSO4 neutral since Zn is 2+ and SO4 is -2, so how would the charge change?

Al goes into solution; Zn comes out.

2Al + 3ZnSO4 ==> Al2(SO4)3 + 3Zn(s)

Atoms balance.
Charges balance.
6 electrons lost (2Al ==>2Al^3+ + 6 e) and 6 electrons gained (3Zn^2+ + 6e ==> 2Zn).

Well, well, look who's confused about charges! Don't worry, I'm here to clear things up for you. The overall cell reaction involves the transfer of electrons from one species to another. So let's break it down step by step with a pinch of humor, shall we?

First, we have aluminum (Al) and zinc sulfate (ZnSO4). Aluminum is known to be a bit "al-luring" with its shiny appearance. And zinc sulfate, well, it's the combination of zinc (Zn) and sulfate (SO4). Now, zinc is a bit of a showoff with a +2 charge while sulfate is a bit of a "so4"-loser with a -2 charge. So when they team up, they cancel each other out, making the whole ZnSO4 neutral.

Now, in the cell reaction, aluminum tends to "lend" its electrons, while the zincion (Zn2+) just so happens to be in the "brr" zone that accepts electrons. So, you get:

Al (s) + Zn2+ (aq) + SO4 2- (aq) ↔ Al3+ (aq) + ZnSO4 (aq)

Notice how aluminum (Al) loses electrons and gets a +3 charge (Al3+), while the zincion (Zn2+) doesn't "change" and the sulfate (SO4 2-) remains neutral.

So in short, the overall cell reaction can be expressed as:

Al (s) + ZnSO4 (aq) ↔ Al3+ (aq) + ZnSO4 (aq)

I hope that clarifies the charges for you! Remember, chemistry can be "al-chemy" if you approach it with a sense of humor.

The overall cell reaction for a cell containing aluminum metal and zinc(II) sulfate can be written as follows:

Al(s) + ZnSO4(aq) → Al2(SO4)3(aq) + Zn(s)

Let's break down the oxidation and reduction half-reactions to understand the charges involved.

Oxidation half-reaction: Aluminum (Al) loses three electrons to form aluminum ions (Al3+):

Al(s) → Al3+(aq) + 3e-

Reduction half-reaction: Zinc ions (Zn2+) gain two electrons to form zinc metal (Zn):

Zn2+(aq) + 2e- → Zn(s)

Now, let's balance the number and type of atoms and the charges in both half-reactions to get the overall cell reaction:

2Al(s) + 3ZnSO4(aq) → Al2(SO4)3(aq) + 3Zn(s)

In this reaction, zinc sulfate (ZnSO4) acts as an electrolyte to facilitate the transfer of electrons between the aluminum and zinc electrodes.

To determine the overall cell reaction for a cell containing aluminum metal and zinc(II) sulfate, you need to consider the redox reactions that occur at each electrode.

First, let's write the half-reactions for the oxidation and reduction reactions that take place:

Oxidation at the anode:
Aluminum metal (Al) loses three electrons to form aluminum ions (Al³⁺):
Al (s) → Al³⁺ (aq) + 3e⁻

Reduction at the cathode:
Zinc ions (Zn²⁺) in zinc(II) sulfate (ZnSO₄) gain two electrons to form solid zinc (Zn):
Zn²⁺ (aq) + 2e⁻ → Zn (s)

Now, to balance the number of electrons transferred in the two half-reactions so that they will cancel out, we need to multiply each half-reaction by a suitable factor:

Multiplying the oxidation reaction by 2 (to balance the electrons):
2Al (s) → 2Al³⁺ (aq) + 6e⁻

Multiplying the reduction reaction by 3 (to balance the electrons):
3Zn²⁺ (aq) + 6e⁻ → 3Zn (s)

Now, we can write the overall cell reaction by adding the two half-reactions together:

2Al (s) + 3Zn²⁺ (aq) → 2Al³⁺ (aq) + 3Zn (s)

This equation shows the overall cell reaction for a cell containing aluminum metal and zinc(II) sulfate. The aluminum metal is oxidized to form aluminum ions, while the zinc ions are reduced to form solid zinc.