Electroplating of silver

  • Thread starter rogen
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  • #1
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I am quite confusing about the concept of electroplating of silver.
As what I know, the silver anode gives out Ag+ while the Ag+ in the silver nitrate solution deposits on the cathode (e.g. a spoon).

However, why the Ag(s) on the silver anode is preferentially discharged instead of the OH- in silver nitrate solution?

Here are the electrode potentials:
O2(g) + 2 H2O + 4 e <--> 4 OH(aq) +0.4V
Ag+ +  e <--> Ag(s) +0.8V
O2(g) + 4 H+ + 4 e <--> 2 H2O +1.23V

As the process is carried out in alkaline condition, I suppose only the first two equations are involved. From the potential values, the OH- should be preferentially discharged rather than the Ag(s), but the actual situation is the opposite. I am not sure if I have misunderstood anything (e.g. the third equation is involved instead).
 

Answers and Replies

  • #2
Borek
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You are comparing standard potentials, you should compare formal potentials, taking into account exact pH of the solution. Solution can't be alkaline - that would mean precipitation of AgOH, so it has to be neutral (if not slightly acidic). Add to that fact that most electrode reactions involving oxygen are sluggish and require substantial overpotentials to proceed quickly.
 
  • #3
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Thanks for your reply:smile:
So should I use the third equation instead of the first one for comparison?
 
  • #4
Borek
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You should start calculating formal potentials for all reactions (using Nernst equation). After that it can be already obvious what is going on. (Or not - I have not checked. But that's the most obvious line of attack).
 
  • #5
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Thankssssss Borek :cool:
I got the answer!
 

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