Oxidation of metals/halogens by oxygen gas in acidic aqueous solution

I had some confusion about the correct answer to the following question:



> 
> Which substance can be oxidized by $\ce{O2(g)}$ in acidic aqueous solution?
> 
> 
> 


Two of the answer choices were metal cations (usually not oxidized further). The remaining three choices were:



> 
> $\ce{Br2(l)}$, $\ce{Ag(s)}$ and $\ce{Br-(aq)}$.
> 
> 
> 


The standard reduction potentials were given:



> 
> $$
> \begin{align}
> \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
> \ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
> \ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
> \end{align}
> $$
> 
> 
> 


The correct answer given was $\ce{Br2(l)}$, however, this did not make sense to me as the oxidation of bromine would result in a positively charged bromine ion. Am I reasoning correctly that $\ce{O2(g)}$ can oxidize both $\ce{Ag}$ and $\ce{Br-}$ in acidic solution?


Since the value of $E^\circ$ for $\ce{O2(g)}$ is greater than both of the other values, more energy is released when $\ce{O2}$ gains the electrons, and this will result in the oxidation of both $\ce{Ag}$ and $\ce{Br-}$, correct?




As noted in your question, the standard *reduction* potentials are given as such in tables. 



> 
> $$
> \begin{align}
> \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
> \ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
> \ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
> \end{align}
> $$
> 
> 
> 


To get the the standard *oxidation* potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:



> 
> $$
> \begin{align}
> \ce{Ag(s) &→ Ag+ + e- } &\quad E^\circ &= \pu{-0.799 V} \\
> \ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
> \end{align}
> $$
> 
> 
> 


Now there are two possible overall reactions (ignoring the Nernst equation):



> 
> $$
> \begin{align}
> \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
> \ce{Ag(s) &→ Ag+ + e- } &\quad E^\circ &= \pu{-0.799 V} \\
> \ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E\_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
> \end{align}
> $$
> 
> 
> 


and the other overall reaction is 



> 
> $$
> \begin{align}
> \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
> \ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \\
> \ce{O2(g) + 4 H+(aq) + 4 Br-(aq) &→ 2Br2(l) + 2 H2O(l)} &\quad E\_\text{Total}^\circ &= \pu{+1.23 - 1.065 = +0.16 V}
> \end{align}
> $$
> 
> 
> 


Since the EMF for both overall reactions is positive, the reactions should be spontaneous. If the overall EMF was negative then the reaction would proceed in the opposite direction (in other words, swap the products and reactants.) 


Notes:


(1) I don't know why the silver reaction doesn't happen. The EMF for the cell indicates that silver should oxidize in these conditions. 


(2) The EMF's calculated ignore the Nernst equation.


(3) The oxygen half cell reaction is a bit weird. Obviously the oxygen must be dissolved in the aqueous solution to react. I'm assuming that the notation is trying to point out that the dissolved oxygen is in equilibrium with the atmospheric oxygen which has a constant concentration which is inexhaustible. 


 2

The sign of the electrode reduction potential is *invariant*. If reflects the sign of the electrostatic charge of the electrode with respect to the hydrogen electrode.\* Also remember that all electrode potentials are written as reduction these days. This is a convention set by all electrochemists all over the world along with the other conventions. The following points are needed to approach this problem. I would suggest to solve this problem conceptually along with your own reasoning. It will give you a better understanding in future. 


a) Ecell=Ecathode - Eanode > 0 implies a spontaneous reaction. (Eq. 1)


b) Don't change any signs of from tabulated values for Eq. 1. Eq. 1 already takes care of that. In the tables, carefully note that all reduced species appear on the left side and all the oxidized forms appear on the right. Reduction always occurs at the cathode, oxidation at the anode.


c) Ecell < 0, it implies a non-spontaneous reaction. However the reaction is spontaneous in the reverse direction.


Coming to your question. They are asking which substance can be oxidized by oxygen? They have also given you the starting materials/ elements.


Since oxygen is supposed to oxidize Br- and Ag, so it must reduce itself. Note that only reduced forms or the species on the right side of the equations of electrode potentials can be oxidized. Br- and Ag are reduced forms, that is why they were chosen for testing via Eq 1. This implies that we choose Ecathode as the oxygen half cell and test the oxidizable materials.


*For Ag+/Ag half cell, and we treat Ag+/Ag as an anode (where silver metal will be oxidized)* 


Ecell = E(cathode)-Eanode = 1.23 - (+0.799) > 0, this is spontaneous, at least in theory. This number does not tell you how fast the reaction will occur or will it occur or not.


For example, ask the same question, will O2 oxidize elemental hydrogen? Ecell is positive, however if you combine O2 and H2 and seal them in a flask, and wait until the age of retirement, nothing will happen. 


*For Br2/Br- half cell, which we are considering as anode where Br- will be oxidized*


Ecell = 1.23 - (+1.065) > 0, this is also spontaneous.


In short the question is good, but your textbook answer is wrong. O2 in acidic medium, at least thermodynamically, can oxidize Ag to Ag+ and Br- to Br2.




---


\*I don't know if you like history or not, long time ago in the 1950s-60s showing the electrostatic sign of a cell by means of a specially designed electroscope was a standard experiment in physics.


Imagine if we say H2O (l) --> H2O (g) at 100 oC


Does this mean reversing the reaction


H2O(g) --> H2O (l) will be at -100 oC?


One can see logical fallacy in reversing the sign of electrode potentials. Allan J. Bard and other leading electrochemists propose to treat the sign of the electrode potentials as constant. It was Latimer who purported the "oxidation potentials" in his very famous book, however, this school of thought is no longer appreciated by electrochemists.


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