Question

(Use the information in the table below. Show the half reactions and the potentials for each half reaction.)

Determine whether the following redox reaction is spontaneous as written. 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s)

Reduction Potentials at 25°C
Electrode Half-Reaction E0(V)
Mg2+/Mg Mg2+ + 2e– → Mg –2.37
Al3+/Al Al3+ + 3e– → Al –1.66
Zn2+/Zn Zn2+ + 2e– → Zn –0.76
Ag2+/Ag Ag+ + e– → Ag +0.80

Answer 1

This is done just like the Ag/Mg^2+ I did earlier. What is it you don't understand?

Answer 2

I'm not sure how to write out the equation . . . & I don't want to get it wrong.

Answer 3

Here is the earlier problem which I showed was -Ecell which makes it non-spontaneous.

2Ag(s) + Mg2+(aq) → 2Ag+ (aq) + Mg(s) [5 points]
You just pair them up. 2Ag(s) on the left goes with 2Ag^+ on the right.
Then Mg^2+ on the left goes with Mg(s) on the right
Then you pick the Ag^+ +e ==> Ag(s) from the table (it's the reverse in the table so you reverse the sign and make it -0.80). Then you pick the Mg^2+ + 2e ==> Mg(s) from the table to get the -2.37. The Ecell is the sum of the two.
The problem of Al and Zn^2+ is done the same way. So have at it. Pairing them up doesn't look that complicated. Have at it and I'll check it for you.

Answer 4

Is this correct:

2Al(s) + 3Zn2+(aq)  2Al3+(aq) + 3Zn(s)
2Al(s) + 2A13+(ag) E = +1.66
3Zn2+(aq) + 3Zn(s) E = +0.76
2Al(s) + 3Zn2+  2Al3+ + 3Zn(s) E = 2.42
The E cell is positive therefore it is spontaneous

Answer 5

Almost. The balanced equation is correct and the two half reactions are correct. One of the potentials is wrong. Here is the table I've printed.

Al3+/Al Al3+ + 3e– → Al –1.66
Zn2+/Zn Zn2+ + 2e– → Zn –0.76.
You have the Al couple correct. It is listed as Al^3+ to Al which is the reverse of what you want so you correctly changed the sign to +1.66. But the other one you want Zn^2+ to Zn and that's the way the table gives it; therefore, you don't change the sign. but keep it at = -0.76. Ecell is still positive and it will be spontaneous. You didn't great.

Answer 6

It may be worth pointing out that when dealing with these equations, you will ALWAYS change the sign of one because it will always be the reverse of what you want but the other half cell will not change the sign.

Answer 7

Thank you so much! ! 😊

Answer 8

I just reread my post and I said "You didn't great!" Obviously I meant YOU DID GREAT! Good job.

Answer 9

Thank you Sir! Today I'm doing my last Chemistry test, & I have the 2nd semester exam due in about two days. It will be the end of chemistry class & I'm kinda sad about it. 😕 I learned a lot during the class & you helped me a lot also. Thank you! :)

Answer 10

[AI-generated answer — Explain Bot]

To determine whether a redox reaction is spontaneous as written, we need to compare the reduction potentials of the half reactions involved.

The given redox reaction is: 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s)

We can break down this reaction into two half reactions: one for the reduction and one for the oxidation.

Reduction half reaction: Al3+ + 3e– → Al
Oxidation half reaction: Zn → Zn2+ + 2e–

Now, let's find the reduction potentials for these half reactions:

The reduction potential for the Al3+/Al half reaction is -1.66 V.
The reduction potential for the Zn2+/Zn half reaction is -0.76 V.

To determine if the overall reaction is spontaneous, we need to compare the reduction potentials of the two half reactions. The more positive reduction potential will be reduced, while the more negative reduction potential will be oxidized.

In this case, the reduction potential for the Al3+/Al half reaction is more negative than the reduction potential for the Zn2+/Zn half reaction. Therefore, Al3+ will not be reduced, and Zn will be oxidized.

Since the more positive reduction potential is associated with the oxidation of Zn, it means that the overall reaction is spontaneous as written.

In summary, the given redox reaction 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s) is spontaneous as written based on the comparison of reduction potentials.