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

Solution :

Step 1: Identification of Electrolytic Species

In an aqueous solution of silver nitrate ($AgNO_{3}$), the following ions are present:

  • From the electrolyte: $Ag^{+}$ and $NO_{3}^{-}$
  • From the solvent (water): $H^{+}$ and $OH^{-}$

[Per the principle of electrolytic dissociation in aqueous media].

Step 2: Analysis of Competitive Discharge at the Cathode

At the cathode, reduction occurs. The competing reactions are:

  1. $Ag^{+}(aq) + e^{-} \rightarrow Ag(s) \quad (E^{\circ} = +0.80 \, V)$
  2. $2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq) \quad (E^{\circ} = -0.83 \, V)$

Since the standard electrode potential ($E^{\circ}$) for the reduction of $Ag^{+}$ is significantly higher (more positive) than that of water, $Ag^{+}$ ions will be preferentially reduced at the cathode [Per the Electrochemical Series hierarchy].

Step 3: Analysis of Competitive Dissolution/Discharge at the Anode

Because the electrodes are specified as Silver (Ag), the anode is active. In cells with active metallic electrodes, the electrode material itself undergoes oxidation if its oxidation potential is higher than that of the anions or water.

The competing reactions are:

  1. $Ag(s) \rightarrow Ag^{+}(aq) + e^{-} \quad (E^{\circ}_{ox} = -0.80 \, V)$
  2. $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-} \quad (E^{\circ}_{ox} = -1.23 \, V)$
  3. $NO_{3}^{-}(aq) \rightarrow \text{No oxidation at these potentials.}$

Since the oxidation potential of silver ($+0.80 \, V$) is greater than that of water ($-1.23 \, V$), the silver anode will dissolve into the solution [Per the Selective Discharge Theory for active electrodes].

Visual Representation of the Electrolytic Cell

Cathode Anode Ag⁺ → Ag(s) Ag(s) → Ag⁺ AgNO₃(aq)

Step 4: Conclusion of Products

Electrode Product Reaction
Cathode Silver metal ($Ag$) $Ag^{+} + e^{-} \rightarrow Ag$
Anode Silver ions ($Ag^{+}$) $Ag \rightarrow Ag^{+} + e^{-}$

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


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