Unit 4 questions thread

[soccerboi]

Hi guys, my first question for unit 4:

Would the equilibrium expression still be the same if the equation was reversed?

e.g wA+xB <--> yC

k=[C]^y/([A]^wX[B ]^x)

but if the eqn was reversed i.e yC <--> wA + xB

Would the equilibrium expression still be  k=[C]^y/([A]^wX[B ]^x)?

[FlorianK]

Let's call the new K for equilibrium Kn.
Kn=K^-1

And you multiply [A]^q with ^x not add them

[Graphite]

Or you can start from scratch, to find the equilibrium constant by
Which shows how the new K is related to the initial K

[soccerboi]

Thanks guys, i forgot that when u reverse the eqn, u simply take the reciprocal of the original equilibrium expression. And yeh the denominator should have been multiply, that was a typo.

[soccerboi]

Lead acid accumulator question

From neap study guide
Q) Write the equation  occurring at the cathode while the cell is recharging.
A) PbSO₄(s)+H⁺+2e^---->Pb(s)+HSO₄^-

From checkpoints
Q) Give the equation of the reaction that occurs at the negative electrode when the accumulator is being recharged.
A) PbSO₄(s)+2e^---->Pb(s)+SO₄^2-

The first question asks for at the cathode and the second question asks for the negative electrode(anode) so why are the equations the same? Which one is correct?

[soccerboi]

In a particular galvanic cell, as current is drawn the cell reaction occurring is
Ag₂O(s) + Zn(s) ---> 2Ag(s) + Zn(OH₂)(s)

Give the eqn for the reaction occurring at the anode and cathode.

Could someone please show me how to split it into the two separate eqns. I don't know how to do it correctly.

I got:
Zn + 2H2O ---> Zn(OH)2 +2H⁺+2e^-
Ag2O +2H⁺ +2e^- ---> 2Ag + H2O

but answers have it as:
Zn + 2OH- --->Zn(OH)2 +2e^-
Ag2O +H2O +2e^- ---> 2Ag +2OH^-

[soccerboi]

In galvanic cells, why must the oxidant and reductant not come into contact? What happens if they do contact?

[Jenny_2108]

Why does your school study so fast? Mine has just finished AOS1

[soccerboi]

Why does your school study so fast? Mine has just finished AOS1

Nah we're not fast, we skipped ch 23,24,25 and will come back to it. They wanted to get the topics which require more conceptual understanding out of the way first.

[golden]

In galvanic cells, why must the oxidant and reductant not come into contact? What happens if they do contact?

They must be kept separate so that they do not react directly - that would use up all your energy source at once, and potentially generate a lot of heat, which all together is a waste (because they are usually used as an energy source for a long time).

[soccerboi]

Probably a simple question but i donno why it is:

Why can't Al be deposited from aqeous solutions of Al3+ ?

Thanks

[jaydee]

Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.

[soccerboi]

Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.

Oh ok thank you, but how do you know which water eqn to look at? I can see 4 on the electrochemical series that contain water.

[Jenny_2108]

Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.

Oh ok thank you, but how do you know which water eqn to look at? I can see 4 on the electrochemical series that contain water.

Depending on the question

[jaydee]

oxidant you usually look at -0.83V and for reductants you would usually look at +1.23V (as a generalisation). yet to fail on me!