Question 3

[guest]

Can you please explain how you arrived at 17 for the answer? Thank you!

[admin]

OK, I would start at the top of the table with Florine (E = 2.87) so basically any voltage with 0.87 or under would have a voltage difference of at lease 2 V with F. When you look down the table, you should see 8 voltaic cells.

Next down the line: Cl at 1.36 V meaning that the cell would have to be - 0.64 or less to have a 2 V difference. The table shows only 2 results. Now we have a total of 10 results.

Next down the line: Br at 1.09, thus we need less that - 0.91 and there is only 1 result. And then it gets faster! Just one result for the next few until you are stuck with 17!

[mkclaiborn1605]

When you are choosing 2 half reactions to make a voltaic cell, if you choose F2 as one of the half reactions, it is the substance that will be reduced (as it has the higher reduction potential). Why then do you not change the signs of the other half reactions since they are being oxidized and not reduced? If you do this for all of the possible pairings, you get a number less than 17.

[mcat_premed3832]

Yes, you definitely change the signs of the oxidized reactions and that's how you get to 17.

For example, start with F2 with a voltage of 2.87. By looking at the table, you will see that Ag and below (a total of 8 half reactions) will add to at least a total cell voltage of 2 V AFTER changing the sign and adding it to 2.87.

Then you continue as described in the Explanation.

[jsfkt78927]

awesome post admin....not its the same as the description. Okay ive stared at these numbers enough, im not seeing where the extra 6 come from?

anyone?

[dwstar183700]

So I see this as:

with F2 -> everything from 0.80V (Ag+) to -2.71V (Na+) (8 total)

with Cl2 -> -2.71 and -0.76 (2 total)

Br2 -> -2.71 (1 total)

Ag+ -> -2.71 (1 total)

Basically, from Fe3+ down to Ni2+, they all can make a voltaic cell over 2V with just Na+. Therefore you get 8 from F2, 2 from Cl2 and from Br2 to Ni2+, you get 7, making a total of 17.