Question

Using the following portion of the activity series for oxidation half reactions
K(s) → K+(aq) + e-
Al(s) → Al3+(aq) + 3e-
Fe(s) → Fe2+(aq) + 2e-
Sn(s) → Sn2+(aq) + 2e-
determine which reaction will occur.

K(s) with Sn2+(aq)
Al3+(aq) with Fe(s)
K+(aq) with Fe2+(aq)
Al(s) with Sn(s)

Answer 1

@uri @hba

Answer 2

@somy can u help w/ this>

Answer 3

no idea
im guessing B

Answer 4

k, thanks

Answer 5

@aaronq

Answer 6

i don't know how to think of this question

Answer 7

I'll try to give you a start. You have a list of relative strengths of oxidizing and reducing agents. The ones higher up on your list are the strongest reducing agents. (They will have the greatest tendency to become oxidized.) The ones closer to the bottom of your list are the strongest oxidizing agents. (They have the greater attraction for electrons. They will also have the greatest tendency to become reduced.)

Your half reactions are shown as oxidations. Some tables reverse this, showing the strongest oxidations on top. The half reactions are written as reductions on those tables. Note this if you use different sources. It can be very confusing.

For the reaction to be spontaneous, the reducing agent has to be higher up your list than the oxidizing agent. Use this idea to compare the locations of the metals and ions in your lists to determine if the reaction will occur spontaneously.

It will take a while to get the hang of this. For example, comparing the positins of K(s) with Sn2+(aq) in your first example, potassium is higher up your list than the stannous ion. The stannous ion is the strongest oxidizing agent. It will tend to be reduced. So, reverse the half reaction for Sn.

Now you have the two half reactions written correctly, one for an oxidation process, and the other for a reduction. So now, will the reaction be spontaneous? One way to tell is to find E0, if you have the standard 1 molar SATP E0 values in a table. Another way is by comparing their reduction potentials. The stannous ion (half reaction is now written as a reduction) is the strongest oxidizing agent, and potassium is the strongest reducing agent, so the reaction will be spontaneous.

Try it for the others yourself.