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Q2.18 (ii):
Predict the products of electrolysis in each of the following: (ii) An aqueous solution of (AgNO_{3}) with platinum electrodes.

Solution :

Step 1: Identification of Ionic Species in Aqueous Solution

In an aqueous solution of silver nitrate ($AgNO_{3}$), the substance dissociates completely into its constituent ions. Water ($H_{2}O$) also contributes to the ionic population due to auto-ionization.

  • From $AgNO_{3}$: $Ag^{+}_{(aq)}$, $NO_{3}^{-}_{(aq)}$
  • From $H_{2}O$: $H^{+}_{(aq)}$, $OH^{-}_{(aq)}$

[Per the Arrhenius theory of electrolytic dissociation]

Step 2: Analysis of Cathodic Reduction

At the cathode, reduction occurs. We compare the standard reduction potentials ($E^{\circ}$) of the competing cations ($Ag^{+}$ and $H^{+}$):

  • $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$; $E^{\circ} = +0.80\,V$
  • $2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}$; $E^{\circ} = 0.00\,V$

Since the reduction potential of $Ag^{+}$ is significantly higher than that of $H^{+}$, $Ag^{+}$ ions are preferentially reduced at the cathode [Per the Nernstian principle of competitive reduction].

Step 3: Analysis of Anodic Oxidation

At the inert platinum (Pt) anode, we compare the oxidation potentials of the competing anions ($NO_{3}^{-}$ and $OH^{-}$/H_{2}O):

  • $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}$; $E^{\circ}_{ox} = -1.23\,V$
  • $2NO_{3}^{-}_{(aq)} \rightarrow N_{2}O_{6(g)} + 2e^{-}$; $E^{\circ}_{ox}$ is highly negative (thermodynamically unfavorable)

Water molecules are oxidized in preference to nitrate ions because the oxidation potential of water is higher (more favorable) than the oxidation potential of $NO_{3}^{-}$, and nitrate ions are kinetically stable [Per the Electrochemical Series hierarchy].

Step 4: Summary of Electrolytic Reactions

Location Half-Reaction Product
Cathode $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$ Silver metal ($Ag$)
Anode $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}$ Oxygen gas ($O_{2}$)
Anode (-) Cathode (+) $H_{2}O \rightarrow O_{2}$ $Ag^{+} \rightarrow Ag$

Final Solution: The product at the cathode is metallic silver ($Ag$) and the product at the anode is oxygen gas ($O_{2}$).


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