Question

I'm having some issues calculating The Gibbs Energy Change for a given reaction where I look up table values-but, once again, one of my table values is just...missing.

Answer 1

@JFraser , my setup is that I have
\[\rm 3Cu^{2+}(aq)+2Cr(s) \rightarrow 3Cu(s)+2Cr^{3+}(aq)\]

Answer 2

The table values that I can find are solely for the copper ion's standard formation; \[ \triangle G^{\circ}\rm \big(Cu^{2+}\big)=65.5 \ kJ/mol\]

Answer 3

But there just...isn't one in my book for the Chromium ion.

Answer 4

Here's the Appendix area from my book that should have a Chromium ion, but it doesn't I'm not crazy. http://i.imgur.com/XzmNnHs.png

Answer 5

can you use the cell voltage you calculated from the previous problem and use \[\Delta G = -nFE\]?

Answer 6

you know the voltage of this cell is 1.08V, so just plug that in for E

Answer 7

Back, sorry about that. One second.

Answer 8

Oh! Okay, that makes more sense.

Answer 9

like I mentioned earlier, you can solve for \(\Delta G\) lots of ways, you don't HAVE to use thermo tables

Answer 10

\[\triangle G^{\circ}=-nFE^{\circ}=\]

Alright, now I just need to find n....do I have to do the whole half-reaction process in this instance to find it?

Answer 11

you can, but it might be quicker to look at the charges of the reactants and products

Answer 12

I'm thinking n = 6?

Answer 13

the copper goes from +2 to 0 (2e- difference), but the chromium goes from 0 to +3

Answer 14

it is 6

Answer 15

\[\triangle G^{\circ}=-nFE^{\circ}=(-6)(96,500)(1.08)=625,320 \rm \ Joules.\]

Answer 16

(Whoops, forgot the negative sign at the end.)

Answer 17

it is

Answer 18

Alright, cool, I got the right answer. Thank you very much for that plug-in suggestion, made things way less of a headache.

Answer 19

you want to be very careful about throwing a unit onto the end of a problem, without carrying the units THROUGH inside the problem

Answer 20

I'll try to watch it in the future. I'm just trying to cover ground quick, because my test is in three hours. Figuring out how to delegate what amount of time to which concepts.

Thank you, again.