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Chemistry Question on Redox Reactions And Electrode Processes

Predict the products of electrolysis in each of the following:

  1. An aqueous solution of AgNO3AgNO_3 with silver electrodes
  2. An aqueous solution AgNO3AgNO_3 with platinum electrodes
  3. A dilute solution of H2SO4H_2SO_4 with platinum electrodes
  4. An aqueous solution of CuCl2CuCl_2 with platinum electrodes.
Answer

(i) AgNO3AgNO_3 ionizes in aqueous solutions to form Ag+Ag ^+ and NO3−NO_3^- ions.
On electrolysis, either Ag+Ag ^+ ions or H2OH_2O molecules can be reduced at the cathode. But the reduction potential of Ag+Ag ^+ ions is higher than that of H2OH_2O.

Ag(aq)++e−→Ag(s);E0=+0.80VAg^+_{(aq)} +e^−\rightarrow Ag_{(s)};E^0=+0.80 V

2H2O(l)+2e−→H2(g)+2OH(aq)−;E0=−0.83V2H_2O_{(l)}+2e^−\rightarrow H_2 (g)+2OH^−_{(aq)};E^0=−0.83 V
Hence, Ag+Ag ^+ ions are reduced at the cathode.
Similarly, AgAg metal or H2OH_2O molecules can be oxidized at the anode. But the oxidation potential of AgAg is higher than that of H2OH_2O molecules.

Ag(s)→Ag(aq)++e−;E0=−0.80VAg_{(s)}\rightarrow Ag^+_{(aq)}+e^−;E^0=−0.80 V

2H2O(l)→O2(g)+4H(aq)++4e−;E0=−1.23V2H_2O_{(l)}\rightarrow O_2(g)+4H^+_{(aq)}+4e^-;E^0=-1.23 V

Therefore, AgAg metal gets oxidized at the anode.

(ii) PtPt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2O_2. At the cathode, Ag+Ag ^+ ions are reduced and get deposited.

(iii) H2SO4H_2SO_4 ionizes in aqueous solutions to give H+H ^+ and SO42−SO_4^{2-} ions
H2SO4(aq)→2H(aq)++SO42−(aq)H_2SO_4(aq) \rightarrow 2H^+_{(aq)}+SO^{2−}_4(aq)
On electrolysis, either of H+H ^+ ions or H2OH_2O molecules can get reduced at the cathode. But the reduction potential of H+H ^+ ions is higher than that of H2OH_2O molecules.

2H(aq)++2e−→H2(g);E0=0.0V2H^+_{(aq)}+2e^−\rightarrow H_2( g);E^0=0.0 V

2H2O(aq)+2e−→H2(g)+2OH(aq)−;E0=0.83V2H_2O_{(aq)}+2e^−\rightarrow H_2( g)+2OH^−_{(aq)};E^0=0.83 V

Hence, at the cathode, H+H ^+ ions are reduced to liberate H2H_2 gas.
On the other hand, at the anode, either of SO42−SO_4^{2-}ions or H2OH_2O molecules can get oxidized. But the oxidation of SO42−SO_4^{2-} involves breaking of more bonds than that of H2OH_2O molecules.

Hence, SO42−SO_4^{2-} ions have a lower oxidation potential than H2OH_2O.
Thus, H2OH_2O is oxidized at the anode to liberate O2O_2 molecules

(iv) In aqueous solutions, CuCl2CuCl_2 ionizes to give Cu2+Cu ^{2+} and Cl−Cl ^- ions as:
CuCl2(aq)→Cu(aq)2++2Cl−(aq)CuCl_2(aq)\rightarrow Cu^{2+}_{(aq)}+2Cl^−(aq)
On electrolysis, either of Cu2+Cu ^{2+} ions or H2OH_2O molecules can get reduced at the cathode. But the reduction potential of Cu2+Cu ^{2+}is more than that of H2OH_2O molecules.

Cu2+(aq)+2e−→Cu(aq);E0=+0.34VCu^ 2+_{(aq)}+2e^- \rightarrow Cu_{(aq)};E^0=+0.34 V
H2O(l)+2e−→H2(g)+2OH−;E0=−0.83VH_2O_{(l)}+2e^- \rightarrow H_2(g) + 2OH^- ;E^0=−0.83V

Hence, Cu2+Cu ^{2+} ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl−Cl ^- or H2OH_2O is oxidized. The oxidation potential of H2OH_2O is higher than that of Cl−Cl^- .

2Cl−(aq)→Cl2(g)+2e−;E0=−1.36V2Cl^-(aq)\rightarrow Cl_2(g)+2e^-;E^0=-1.36V

2H2O(l)→O2(g)+4H(aq)++4e−;E0=−1.23V2H_2O_{(l)}\rightarrow O_2(g)+4H^+_{(aq)}+4e^−;E^0=−1.23 V
But oxidation of H2OH_2O molecules occurs at a lower electrode potential than that of Cl−Cl ^- ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl−Cl ^- ions are oxidized at the anode to liberate Cl2Cl_2 gas.