Exam question solutions seem wrong?

[PB]

the engage unit 4 exam (the one before the 2013 one) second last question.
http://engageeducation.org.au/practice-exams/chemistry

i thought that since Cu2+ is the stronger oxidant it would have been reduced (thus reacts at the cathode?)
solutions say other wise :PP

Thanks in advance for helping! +1

[SocialRhubarb]

It specifies that it is an electrolytic cell, not a galvanic cell. We can assume that electrolysis is occurring, not a spontaneous redox reaction.

Hence, copper metal is oxidised at the anode and iron metal is reduced at the cathode, because the reaction which is occurring must be electrolysis.

EDIT: I don't mean to hijack your thread, but I think the answer to question 8 multiple choice is incorrect?

[PB]

thanks for replying rhubarb, however I am not quite sure I understand you...I am aware that it is an electrolytic cell, which is why the strongest oxidant should react with the strongest reductant right? so the strongest oxidant(according to the electrochemical series) is Cu2+ and thus should be reduced at the cathode?

and yes, I did question the solution of Multchoice 8...I initially thought that by reducing the [H+] through dilution, the [OH-] would automatically increase as Kw has to remain constant at 10^-14.  However, i think what they mean is that by adding water, EVERYTHING in the mixture is diluted. because I mean, you haven't added any H+ ions or OH- ions in, so the same amount of ions divided by a larger volume would equal a smaller concentration (C=n/V).

But I don't know...hopefully some of the big guns of chem can help us out!

[SocialRhubarb]

In chemistry and manufacturing, electrolysis is a method of using a direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction.

- Wikipedia

You may be confusing a galvanic cell with an electrolytic cell.

Electrolysis has to be a reaction which is 'non-spontaneous', one which would not occur naturally, without the input of electrical energy. If you add iron metal to a solution with copper (II) ions, the two will spontaneously react to form iron (II) ions and copper metal. Hence, this cell is not electrolytic, but galvanic, since the two have spontaneously reacted. Therefore, to have an electrolytic cell, we need to have iron (II) ions reacting with copper metal.

Also, you are adding OH- ions in. Water contains hydroxide ions, at a concentration of 10^-7 M at 25 degrees Celsius. Normally this is low enough to ignore, but we're adding it to an acid which is going to have an even lower concentration of hydroxide. So the concentration of hydroxide should increase. You're right in saying that the Kw has to remain constant (though not necessarily at 10^-14), and adding water shouldn't change the constant.

EDIT: grammar.

[PB]

Ah cheers, rhubarb so what you are saying is the the gradient on the electrical series basically has to be positive (from left to right) so like, you need electrical energy to climb up that hill?

And yeh, I think you may be right with the mult choice too. Hopefully nliu or something could help us out

[lzxnl]

Yep, it says electrolytic cell; you start off with copper metal and iron(II) ions and reverse the conventional reaction to form copper ions and iron metal.

[PB]

Cheers for replying nliu, but I think we are most concerned about Multiple Choice Q8!

[lzxnl]

- Wikipedia

You may be confusing a galvanic cell with an electrolytic cell.

Electrolysis has to be a reaction which is 'non-spontaneous', one which would not occur naturally, without the input of electrical energy. If you add iron metal to a solution with copper (II) ions, the two will spontaneously react to form iron (II) ions and copper metal. Hence, this cell is not electrolytic, but galvanic, since the two have spontaneously reacted. Therefore, to have an electrolytic cell, we need to have iron (II) ions reacting with copper metal.

Also, you are adding OH- ions in. Water contains hydroxide ions, at a concentration of 10^-7 M at 25 degrees Celsius. Normally this is low enough to ignore, but we're adding it to an acid which is going to have an even lower concentration of hydroxide. So the concentration of hydroxide should increase. You're right in saying that the Kw has to remain constant (though not necessarily at 10^-14), and adding water shouldn't change the constant.

EDIT: grammar.

Yep, I agree with Social Rhubarb on this one. Your 0.5 M HCl has roughly 2*10^-15 M OH-, but you're adding 10^-7 M hydroxide. That's actually VERY significant.

[PB]

Ok, ok I concede Thanks SocialRhubarb and nliu!

[Patches]

That acid dilution question is the same as multiple choice question 8 on the VCAA 2010 Exam 2 if you want a proper solution.

[lzxnl]

That acid dilution question is the same as multiple choice question 8 on the VCAA 2010 Exam 2 if you want a proper solution.

So our solutions aren't good enough for you?
Kidding

[PB]

Thanks patches!