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

Solution :

Step 1: Identification of Ionic Species in Solution

In an aqueous solution of silver nitrate ($AgNO_{3}$), the substance dissociates completely according to the following equation: $AgNO_{3}(aq) \rightarrow Ag^{+}(aq) + NO_{3}^{-}(aq)$

Since it is an aqueous solution, water ($H_{2}O$) molecules are also present, which can undergo oxidation or reduction [Per the Nernst Theory of Electrolytic Dissociation].

Step 2: Analysis of Potential Electrode Reactions

At the electrodes, the species with higher reduction potential (for cathode) or lower oxidation potential (for anode) will preferentially react. Because silver electrodes are used, the electrodes are active, meaning they participate in the reaction.

At the Cathode (Reduction):

  • Reduction of $Ag^{+}$ ions: $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$ ($E^{\circ} = +0.80 \, V$)
  • Reduction of $H_{2}O$: $2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq)$ ($E^{\circ} = -0.83 \, V$)

Since the reduction potential of $Ag^{+}$ is significantly higher than that of $H_{2}O$ [Per the Electrochemical Series], $Ag^{+}$ ions are preferentially reduced to metallic silver.

At the Anode (Oxidation):

  • Oxidation of $Ag$ electrode: $Ag(s) \rightarrow Ag^{+}(aq) + e^{-}$ ($E^{\circ}_{ox} = -0.80 \, V$)
  • Oxidation of $NO_{3}^{-}$ ions: $2NO_{3}^{-} \rightarrow 2NO_{2} + O_{2} + 2e^{-}$ ($E^{\circ}_{ox} \approx -2.0 \, V$)
  • Oxidation of $H_{2}O$: $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$ ($E^{\circ}_{ox} = -1.23 \, V$)

In the presence of an active silver anode, the silver metal itself is more easily oxidized than the water or the nitrate ions [Per the Principle of Preferential Discharge]. Therefore, the anode undergoes dissolution.

Visual Representation of the Electrolytic Cell

Anode (Ag) Cathode (Ag) Ag(s) → Ag⁺(aq) + e⁻ Ag⁺(aq) + e⁻ → Ag(s)

Step 3: Synthesis of Final Products

As the current flows through the circuit:

  1. The silver anode dissolves into the solution: $Ag(s) \rightarrow Ag^{+}(aq) + e^{-}$.
  2. The silver ions from the solution (and those produced at the anode) migrate to the cathode.
  3. The silver ions at the cathode gain electrons and deposit as metallic silver: $Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$.

Effectively, the net result is the transfer of silver from the anode to the cathode.

Final Solution: The product at the cathode is metallic silver ($Ag$), and the product at the anode is the dissolution of the silver electrode (formation of $Ag^{+}$ ions in the solution).


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