Write chemical equations to represent the cathode, anode, and net cell reactions.
1) Cd | Cd(NO3)2 || Ag(NO3)2 | Ag
Reduction at Cathode: Cd2+ + 2e- ---> Cd
Oxidation at Anode: [Ag ---> Ag+ + e-] x2
Cd2+ + 2Ag ---> Cd + 2Ag2+
2) Pt | IO3-,H+ || Zn2+ | Zn
Reduction at Cathode: 2IO3- + 12H+ + 10e- ---> I2 + 6H2O
Oxidation at Anode: [Zn ---> Zn2+ + 2e-] x 5
2IO3- + 12H+ + Zn ---> I2 + 6H2O + Zn2+
To derive the chemical equations for the cathode, anode, and net cell reactions, you need to identify the half-reactions occurring at each electrode.
1) For the given cell:
Cathode: Cd | Cd(NO3)2
Anode: Ag(NO3)2 | Ag
To represent the reduction reaction at the cathode, you should identify the reduced species and its charges. In this case, Cd2+ is reduced to Cd. The equation is:
Cd2+ + 2e- -> Cd
Next, you identify the oxidation reaction at the anode. The species being oxidized is Ag, which loses electrons to form Ag+ ions. Multiply the oxidation half-reaction by 2 to balance the electrons:
[Ag -> Ag+ + e-] x2
Finally, to write the net cell reaction, combine the two half-reactions and cancel out any common species:
Cd2+ + 2Ag -> Cd + 2Ag2+
Therefore, the net cell reaction for the given cell is Cd2+ + 2Ag -> Cd + 2Ag2+.
2) For the second cell:
Cathode: Pt | IO3-, H+
Anode: Zn2+ | Zn
The reduction reaction at the cathode involves IO3-. In an acidic solution, IO3- is converted to I2. The equation is:
2IO3- + 12H+ + 10e- -> I2 + 6H2O
The oxidation reaction at the anode involves Zn. It loses electrons to form Zn2+ ions. The equation is:
[Zn -> Zn2+ + 2e-] x5
To write the net cell reaction, combine the half-reactions and cancel out any common species:
2IO3- + 12H+ + Zn -> I2 + 6H2O + Zn2+
Therefore, the net cell reaction for the given cell is 2IO3- + 12H+ + Zn -> I2 + 6H2O + Zn2+.