I read that the electrolytic refining of Silver is done using dilute silver nitrate as an electrolyte and a small percentage of free $\ce{HNO3}$. The following reactions are possible at the anode:

$$\ce{Ag ->Ag+ + e-}$$ $$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does $\ce{Ag}$ of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

  • 2
    $\begingroup$ The second reaction has E° = + 0.4 V, if the concentration of OH- is 1 M. Please apply Nernst equation to this electrode. $\endgroup$ – Maurice May 29 at 7:44
  • 1
    $\begingroup$ Oxidation always occurs at anode. I think you meant the reverse reaction for Ag $\endgroup$ – user600016 May 29 at 8:26
  • $\begingroup$ @Maurice The value changes with concentration of the electrolyte..so that explains why Ag is oxidised . $\endgroup$ – Chopin May 29 at 10:53
  • $\begingroup$ You have mistaken reactions. Note that: $$\ce{Ag+ + e- <=> Ag} \quad E^\circ = \pu{0.7996 V}$$. $\endgroup$ – Mathew Mahindaratne May 29 at 12:08
  • 1
    $\begingroup$ @Mathew Mahindaratne I wrote down the equations like they would happen at the anode but specified the values of the reduction potential.(reverse reaction) $\endgroup$ – Chopin May 29 at 12:47

Electrolysis or electrolytic refining is a technique used for extraction as well as purification of metals obtained by other refining methods. In the electrolytic refining process, a block or a strip of impure metal is used as the anode, and a thin sheet of pure metal is used as the cathode. Both the cathode and anode are dipped in an electrolytic cell containing an aqueous solution of the metal salt. Upon the application of direct electric current of a suitable voltage, pure metal is deposited at the cathode (reduction) by the dissolution of impure metal at the anode (oxidation).

Since electrolytic refining id done in aqueous medium, the key to successful refining is the oxidation of the metal should be more facile than the oxidation of water (see the standard oxidation potentials below). Hence the solid metal would dissolve into the solution as cataions, leaving behind many of the impurities, which are inactive to electrolysis. The possible anodic oxidation reactions would be:

$$\ce{Ag <=> Ag+ + e-} \quad E_\mathrm{ox}^\circ = \pu{-0.7996 V} \tag1$$ $$\ce{4H2O <=> O2 + 4H+ + 4e- } \quad E_\mathrm{ox}^\circ = \pu{-1.229 V} \tag2$$

When value of $E_\mathrm{ox}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{ox}^\circ \ (\ce{H2O})$, it is safe to say that $\ce{Ag}$ oxidation at anode is more spontaneous (less negative value) than that of $\ce{H2O}$ (greater negative value) at anode. As a result, impure $\ce{Ag}$ anode will be slowly dissolved into solution as $\ce{Ag+}$ ins as direct current applied.

Similarly, the possible cathodic reduction reactions would be:

$$\ce{Ag+ + e- <=> Ag} \quad E_\mathrm{red}^\circ = \pu{0.7996 V} \tag3$$ $$\ce{ 2H2O + 2e- <=> H2 + 2OH-} \quad E_\mathrm{red}^\circ = \pu{-0.8277 V} \tag4$$

Since the refining solution is dilute $\ce{HNO3}$ to provide the counter-ion for $\ce{Ag+}$ cation, we should consider possibility of another reduction reaction at cathode:

$$\ce{NO3- + 3H+ + 2e- <=> HNO2 + H2O} \quad E_\mathrm{red}^\circ = \pu{0.934 V} \tag5$$ $$\text{And, } \ce{NO3- + 4H+ + 3e- <=> NO + 2H2O} \quad E_\mathrm{red}^\circ = \pu{0.957 V} \tag6$$

When the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{red}^\circ \ (\ce{H2O})$, it is clear that $\ce{Ag}$ reduction at cathode is much more prominent (high positive value) than that of $\ce{H2O}$ (greater negative value) at cathode. However, the competition reduction would happen as well when comparing the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ with those of $E_\mathrm{red}^\circ \ (\ce{NO3-})$, both of which are slightly higher positive values. However, if you use appropriately dilute nitric acid to begin with (c.f., Nernst equation), you will be able to avoid nitrate reduction at cathode.

| improve this answer | |
  • $\begingroup$ Why is the reaction involving OH- ions at the anode not a possibility? $\endgroup$ – Chopin May 29 at 14:09
  • 1
    $\begingroup$ The solution is acidic, therefore not enough $\ce{OH-}$ ions floating around the solution. But there are plenty of $\ce{H2O}$ there. If you have used $\ce{NaOH}$ instead of $\ce{HNO3}$, that is a possibility. $\endgroup$ – Mathew Mahindaratne May 29 at 14:14
  • $\begingroup$ @ Mathew Mahindaratne . If NaOH is added to a silver nitrate solution, it would produce a precipitate of silver oxide Ag2O, and silver can not be deposited at the cathode any more. $\endgroup$ – Maurice May 31 at 10:17
  • $\begingroup$ The process of purifying silver by anodic oxidation is working if the impurities are made of metals with redox potentials higher than 0.800 V. If the impurity is made of metals with redox potentials lower than 0.800 V, like copper, this impurity will be oxidized before silver, and silver will be precipitated at the bottom of the container. This is what happens when refining copper mixed with noble metals by anodic oxidation. At the anode, copper is dissolved and impurities like silver are deposited at the bottom of the solution. At the cathode, pure copper is obtained. $\endgroup$ – Maurice May 31 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.