Predict the products of electrolysis in each of the following:
An aqueous solution of AgNO3 with silver electrodes
Predict the products of electrolysis in each of the following:
An aqueous solution AgNO3 with platinum electrodes
Predict the products of electrolysis in each of the following;
A dilute solution of H2SO4 with platinum electrodes.
Predict the products of electrolysis in each of the following:
An aqeuous solution of CuCl2 with platinum electrodes.
Predict the products of electrolysis in each of the following:
An aqueous solution of AgNO3 with silver electrodes
Predict the products of electrolysis in each of the following:
An aqueous solution AgNO3 with platinum electrodes
Predict the products of electrolysis in each of the following;
A dilute solution of H2SO4 with platinum electrodes.
Predict the products of electrolysis in each of the following:
An aqeuous solution of CuCl2 with platinum electrodes.
AgNO3 ionizes solutions to form Ag+ and NO−3
On electrolysis, either Ag+ ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.
Ag+(aq)+e−⟶Ag(s);E∘=+0.80 V
2H2O(l)+2e−⟶H2(g)+2OH−(aq);E∘=−0.83 V
Hence, Ag+ ions are reduced at the Cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of A is higher thatn that of H2O molecules.
Ag(s)⟶Ag+(aq)+e−;E∘=−0.80 V
2H2O(l)⟶O2(g)+4H+(aq)+4e−;E∘=−1.23 V
Therefore, Ag metals gets oxidized at the anode.
Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.
H2SO4 ionizes in aqueous solutions to give H+ and SO2−4 ions.
H2SO(aq)⟶2H+(aq);E∘=0.0 V
On electrolysis, either of H+ ions or H2O molecules can get reduced at the Cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.
2H+(aq)+2e−⟶H2(g);E∘=0.0 V
2H2O(aq)+2e−⟶H2(g)+2OH−(aq);E∘=−0.83 V
Hence, at the cathode, H+ ions are reduced to liberate H2 gas.
On the other hand, at the anode, either of SO2−4 ions or H2O molecules can get oxidized.
But the oxidation of SO2−4 ions have a lower oxidation potential than H2O molecules.
Hence, SO2−4 ions have a lower oxidation potential than H2O is oxidized at the anode to liberate O2 molecules.
In aqeuous solutions, CuCl2 ionizes to give Cu2+ and Cl− ions as:
CuCl2(aq)⟶Cu2+(aq)+2Cl−(aq)
On electrolysis, either of Cu2+ ions or H−2O molecules can get reduced at the Cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.
Cu2+(aq)+2e−⟶Cu(aq);E∘=+0.34 V
H2O(l)+2e−⟶H2(g)+2OH−;E∘=−0.083 V
Hence, Cu2+ ions are reduced at the cathode and get deposited.
Similarly, at the anode, wither of Cl− or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl−.
2Cl−(aq)⟶Cl2(g)+2e−;E∘=−1.36V
2H2O(l)⟶O2(g)+4H+(aq)+4e−;E∘=−1.23 V
But oxidation of H2O molecules occur at a lower electrode potential than that of Cl− ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl− ions are oxidized at the anode to liberate Cl2 gas.
AgNO3 ionizes solutions to form Ag+ and NO−3
On electrolysis, either Ag+ ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.
Ag+(aq)+e−⟶Ag(s);E∘=+0.80 V
2H2O(l)+2e−⟶H2(g)+2OH−(aq);E∘=−0.83 V
Hence, Ag+ ions are reduced at the Cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of A is higher thatn that of H2O molecules.
Ag(s)⟶Ag+(aq)+e−;E∘=−0.80 V
2H2O(l)⟶O2(g)+4H+(aq)+4e−;E∘=−1.23 V
Therefore, Ag metals gets oxidized at the anode.
Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.
H2SO4 ionizes in aqueous solutions to give H+ and SO2−4 ions.
H2SO(aq)⟶2H+(aq);E∘=0.0 V
On electrolysis, either of H+ ions or H2O molecules can get reduced at the Cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.
2H+(aq)+2e−⟶H2(g);E∘=0.0 V
2H2O(aq)+2e−⟶H2(g)+2OH−(aq);E∘=−0.83 V
Hence, at the cathode, H+ ions are reduced to liberate H2 gas.
On the other hand, at the anode, either of SO2−4 ions or H2O molecules can get oxidized.
But the oxidation of SO2−4 ions have a lower oxidation potential than H2O molecules.
Hence, SO2−4 ions have a lower oxidation potential than H2O is oxidized at the anode to liberate O2 molecules.
In aqeuous solutions, CuCl2 ionizes to give Cu2+ and Cl− ions as:
CuCl2(aq)⟶Cu2+(aq)+2Cl−(aq)
On electrolysis, either of Cu2+ ions or H−2O molecules can get reduced at the Cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.
Cu2+(aq)+2e−⟶Cu(aq);E∘=+0.34 V
H2O(l)+2e−⟶H2(g)+2OH−;E∘=−0.083 V
Hence, Cu2+ ions are reduced at the cathode and get deposited.
Similarly, at the anode, wither of Cl− or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl−.
2Cl−(aq)⟶Cl2(g)+2e−;E∘=−1.36V
2H2O(l)⟶O2(g)+4H+(aq)+4e−;E∘=−1.23 V
But oxidation of H2O molecules occur at a lower electrode potential than that of Cl− ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl− ions are oxidized at the anode to liberate Cl2 gas.
