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.

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.