# How to use standard reduction potential table

by TheRedDevil18
Tags: potential, reduction, standard, table
 P: 168 Hi, can someone please explain to me how I can identify which is the oxidation half and which is the reduction half of a redox reaction. I have read on the internet that the oxidation half is the highest up the table between the two then why doesn't it work for: (Ca^2+) + 2Br → Ca + Br2, calcium is higher up the table but it is being reduced. Using table 4B. Am I using the table wrong?
 Admin P: 23,597 Where did you got the reaction equation from? Apart from everything else, it is not balanced.
 P: 168 I don't see anything wrong with the balancing, 1 calcium and two bromines on both sides
 Admin P: 23,597 How to use standard reduction potential table What about charge?
 P: 168 Sorry, (Ca^2+) + (2Br^-) → Ca + Br2 Now could you please answer my question
 Admin P: 23,597 Such reaction would not happen. Other than that - assign oxidation numbers to all atoms. Oxidation means ON goes up, reduction means ON goes down.
 P: 168 I know that but how would I know which is oxidation and which is reduction by using a half reaction table
 Admin P: 23,597 I guess now we are getting to the point. Generally speaking the one with the higher standard potential will oxidize the one with the lower standard potential. So for example, we have two half reactions: Fe3+ + e- <-> Fe2+ (E0 = +0.77V) Cu2+ + 2e- <-> Cu(s) (E0 = +0.34V) Fe3+ will oxidize metallic copper to Cu2+.
 P: 383 I just always work it out by keeping in mind that you need ΔG° < 0 for a reaction to proceed spontaneously and that ΔG° = -nFE°. So when you look at a table of standard reduction potentials you need to flip the half reactions so as to get the most positive E° (when added together) which will then give the most negative ΔG°, showing that the reaction proceeds as written (after balancing etc).
P: 168
 Quote by Borek I guess now we are getting to the point. Generally speaking the one with the higher standard potential will oxidize the one with the lower standard potential. So for example, we have two half reactions: Fe3+ + e- <-> Fe2+ (E0 = +0.77V) Cu2+ + 2e- <-> Cu(s) (E0 = +0.34V) Fe3+ will oxidize metallic copper to Cu2+.

(Zn^2+) + 2e → Zn (-0.76)
(Cu^2+) + 2e → Cu (+0.34)

By standard potential you mean the voltages?, because Cu voltage is higher than Zn but yet Zn oxidises Cu
 P: 343 That table is handy, you can quickly see which substances can be combined into working voltaic cells. For you, it shows the calcium related half reaction is -2.87 V. The bromine related one is +1.07 V. For this to be a working voltaic cell, the total cell potential must be positive (=main idea for these types of problems) E° (cell) = E°(cathode) - E°(anode) (> 0 for voltaic) With those 2 numbers from the table the only way that can happen (only way to make a positive difference) is if bromine is the cathode and calcium the anode. +1.07 - (-2.87) = +3.94 V > 0 Therefore bromine is getting reduced and calcium getting oxidized (Reduction and cathode both start with consonants, oxidation and anode both start with vowels, is how I remember). So your equation would actually be Br2 + Ca (s) -> Ca2+ (aq) + 2Br-(aq) This shows bromine gaining electrons and calcium losing them. Now for your new example with the Zn (-0.76) and Cu (+0.34), the only way to get a working voltaic cell is if Zn is the anode and copper the cathode. Otherwise you would get a negative number. Then again for when writing the equation, you just need to show Zn being oxidized (turning into cations) and copper being reduced (going from ions into a solid). Hope that helps.