Determining reduction or oxidation

[GomerPyle]

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I encountered a weird question in the chemistry self assessment (number 45).

In the passage it states that oxygen is bubbled over a silver electrode where it is reduced according to the equations 1 and 2. The electrochemical reaction is completed at the cadmium electrode. The half reaction is given by equation 3.

Equation 1 - O2 + 2h20 + 4e- --> 4OH
Equation 2 - Ag+ + e- --> Ag(s)
Equation 3 - Cd + 2OH --> Cd(OH)2 + 2e


My question is - if oxygen is bubbled over a silver electrode where it is reduced, doesn't that mean that the silver electrode is oxidized since it reduces oxygen? Then if the electrochemical reaction is completed at the cadmium electrode, then cadmium must be reduced (the cathode) since silver is oxidized? According to the answer, the cadmium electrode is the anode because the silver electrode is where reduction takes place, but I don't understand how silver is where reduction takes place because the oxygen is reduced at the silver electrode where it gets bubbled, and something has to reduce the oxygen (which I thought was the silver, which will then need to oxidize).

Also, I noticed the equations were given in their correct oxidation and reduction forms. Can we just go off the given equations and say that ag was reduced and cd was oxidized, or can we not trust the equations since sometimes the MCAT gives us the equations in various forms?

 

[GomerPyle]

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[nOchemallday]

I encountered a weird question in the chemistry self assessment (number 45).

In the passage it states that oxygen is bubbled over a silver electrode where it is reduced according to the equations 1 and 2. The electrochemical reaction is completed at the cadmium electrode. The half reaction is given by equation 3.

Equation 1 - O2 + 2h20 + 4e- --> 4OH
Equation 2 - Ag+ + e- --> Ag(s)
Equation 3 - Cd + 2OH --> Cd(OH)2 + 2e


My question is - if oxygen is bubbled over a silver electrode where it is reduced, doesn't that mean that the silver electrode is oxidized since it reduces oxygen? Then if the electrochemical reaction is completed at the cadmium electrode, then cadmium must be reduced (the cathode) since silver is oxidized? According to the answer, the cadmium electrode is the anode because the silver electrode is where reduction takes place, but I don't understand how silver is where reduction takes place because the oxygen is reduced at the silver electrode where it gets bubbled, and something has to reduce the oxygen (which I thought was the silver, which will then need to oxidize).

Also, I noticed the equations were given in their correct oxidation and reduction forms. Can we just go off the given equations and say that ag was reduced and cd was oxidized, or can we not trust the equations since sometimes the MCAT gives us the equations in various forms?

This took a while to understand; but here's the quick and dirty explanation:

The first equation is the reduction of oxygen.
This equation oxidizes silver at the silver electrode to generate Ag+ ions in solution at the half cell containing the silver electrode.

The Ag half cell is then hooked up to the Cd half cell to "complete" the electrochemical cell. Equations 2 and 3 are 'correct' in this case because Ag+ ions created by reaction 1 are then reduced back to Ag to create the overall electrochemical cell with Cd (which is oxidized to Cd2+).

Reduction of Ag is at the cathode. Oxidation of Cd is at the anode.