1. 3Fe{2+} (aq) = Fe(s) + 2Fe{3+} (aq)
1)Fe2+(aq)+2e-=Fe(s) redox potential -0.44 V
….2Fe2+(aq)=2Fe3+(aq)+2e- redox potential 0.77V but need to amend 2 because 2 moles reacted
so… 1.54V-0.44V=1.10V spontaneous
2) 3Sn4+(aq)+6e-=3Sn2+ redox potential 0.15V (need to amend for 3)
….2Cr=2Cr3+(aq)+6e- redox potential 0.74V mult par
1.48V+0.45V=1.93V spontaneous
3)3Fe=3Fe2+(aq)+6e- redox potential 0.44V (reversal of sign because the reaction was the opposite of number one)(mult by 3)
…2Cr3+(aq)+6e-=2Cr redox potential -0.74V (see half reaction above for sign reversal reasoning)(mult by 2)
1.32V-1.48V = -0.16V non-spontaneous
4)Sn4+(aq)+2e-=Sn2+ redox potential 0.15V
….Fe2+(aq)+2e-= Fe redox potential -0.44V
0.15V-0.44V=-0.29V not spontaneous
5) not properly balanced
the positive redoex potential for the reaction is spontaneous, the negative is not
I will show you thank you for making ONE. You do something. as big as a million. look up the rate Eo for Fe(2+) + 2 e(-) = Fe, you will find it in a table of daily helping potentials in your book. Now look at the Eo rate for Fe(2+) = Fe(3+) + e(-) you have a hard time looking for the one if your book’s table lists “potential help” because it’s not written as an aid. So write it as a helper (rotate it so the electrons are on the left) but then opposite the Eo rate sign you appear in the chart. load both Eo values together. is the Eo for the completed reaction. you wouldn’t want to be acceptable to the kind of outclassed electrons. If the total Eo level obtained is beneficial, then the reaction is spontaneous as written. this is because Delta G = –n FE, so a favorable capacity corresponds to a destructive delta G, is the thermodynamic criterion of spontaneity. there is a necessary loophole in the previous answer posted. nfj13 mentioned that you want to find Delta G values for reactions of type 0.5. Assigning a Delta G to a nil.5 reaction is not thermodynamically significant, so you won’t find them everywhere. Now I see this was posted earlier. Since you’re expecting someone to show you a much less demanding thank you for trying this, I assure you there isn’t.
Divide each into a reduction and an oxidation.
Find the corresponding electrochemical reduction potentials for each. Reverse the oxidation sign and add them. Next, decide on spontaneity based on the sign of the overall potential.
It would be 2.