Electrolysis using electrodes of metals whose ions are in solution

Question

When we perform electrolysis of say an aqueous solution of silver nitrate with silver electrodes, silver ions, having a greater reduction potential than water, accept electrons and are deposited at the cathode. At the anode however, silver just leaves the electrode and enters into its solution.

The anode must just be a metal rod whose sea of electrons is much poorer than what it must be. Why does the silver get oxidized and leave the rod as ions? Water could get oxidized instead and that appears to be a more logical expectation.

Secondly, why does this not happen if we do the same reaction using platinum electrodes? What is preventing the metal from leaving the anode as an ion, moving to the cathode causing gradual thinning of one electrode and thickening of another?

Answer 1

The reason why silver dissolves (or water is oxidized) is well known. The reason why platinum does not dissolve is more elusive. Thanks to Maurice's comment for pointing this out.

Let's assume your solution is 1M in $\ce{AgNO3}$ and 1M in acid ($\ce{H+}$). The potential required to oxidize silver is 0.80 V; the voltage required to split water (oxidize it) is 1.23 V, but there are complications of overvoltage, mostly with hydrogen, but also with oxygen. To sum up: it is easier to grab an electron from silver than to pump another hydrogen ion into solution.

AHA! How about reducing the acid concentration, or not adding any at all? (Of course, the reason for adding the acid was to make the silver dissolve easier.) If the pH is increased so that hydrogen ion concentration is reduced to give a lower oxidation potential than silver, then water will be oxidized, hydrogen ion will be produced, and eventually, silver will become more easily oxidized. This electrolysis will go through a phase where hydrogen and silver are co-deposited, probably making a porous, weak deposit of silver.

For platinum to yield 2 electrons requires 1.188 V. Although this is less than required for oxidizing water, overvoltage effects due to oxygen and hydrogen are apparently important. Although platinum can be electroplated with a platinum anode, apparently the process fails to work well; older methods of platinum plating were kept refreshed with additions of platinum salts rather than relying on dissolution of the anode (Ref 1). More modern methods use a pulsed electropotential, but their success seems to be from a solving difficulties at the cathode rather than getting the anode to dissolve (Ref 2). They deposit a monolayer of platinum attached to a monolayer of hydrogen, then oxidize off the hydrogen and repeat. The electrode potentials strongly suggest that increasing the cell potential, if you had a platinum anode, would result in dissolving platinum metal, but the catalytic reactivity of platinum with respect to $\ce{H2O2}$ suggests that maybe the oxidation reaction occurs in the water layer, and the platinum acts merely as an electronic conductor.

Ref 1. https://www.technology.matthey.com/article/25/1/32-41/

Ref 2. https://www.chemistryworld.com/news/platinum-plating-at-the-flick-of-a-switch-/5720.article

Answer 2

When current flows in an electrolytic cell, the bulk solution has to remain electrically neutral. If one positive "ion" is deposited at the cathode as a neutral atom, another positive ion has to be generated. What is the source of the positive ion? The metallic anode.

Coming to your other question: What if the anode is a non-metal instead of any metal (an extreme case)? Say, it is made of graphite. In that case, water begins to oxidize, and a hydrogen ion is generated.