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+
For #1, have you done the voltage, Ecell?
Is this reaction to occur spontaneously?
If so, isn't what you show as the cell reaction a negative voltage? (The silver ion should be Ag^+.
The second equation isn't multiplied by 5 for Zn.
1) Cd | Cd(NO3)2 || Ag(NO3)2 | Ag
Reduction at Cathode: [Ag+ + e- ---> Ag]x2
Oxidation at Anode: Cd ---> Cd2+ + 2e-
2Ag+ + Cd ---> Ag + Cd2+
Is this right?
2IO3- + 12H+ + 5Zn ---> I2 + 6H2O + 5Zn2+
2Ag+ + Cd ---> 2Ag + Cd2+ *
For the cell reactions to occur spontaneously, you have equations written correctly and oxidized/reduced equations also are correct. You have omitted a 5 Zn on the left of the second cell and 5Zn^+2 on the right. But you showed it multiplied by 5; I assume the omission was just an oversight.
To write the chemical equations representing the cathode, anode, and net cell reactions, you need to identify the substances that undergo reduction (gain electrons) at the cathode and substances that undergo oxidation (lose electrons) at the anode.
1) In the first example, the cathode consists of a cadmium electrode immersed in a solution of Cd(NO3)2, while the anode consists of a silver electrode immersed in a solution of Ag(NO3)2. The reduction reaction at the cathode is the reduction of Cd2+ ions to form solid Cd:
Cd2+ + 2e- ---> Cd
The oxidation reaction at the anode involves the oxidation of Ag atoms to form Ag+ ions:
Ag ---> Ag+ + e-
Since the oxidation half-reaction involves the loss of one electron, we need to multiply it by 2 to balance the number of electrons transferred in the whole cell reaction:
2Ag ---> 2Ag+ + 2e-
To write the net cell reaction, we can combine the reduction and oxidation half-reactions and make sure the electrons cancel out:
Cd2+ + 2Ag ---> Cd + 2Ag2+
2) In the second example, the cathode consists of a platinum electrode immersed in a solution containing IO3- ions and H+ ions, while the anode consists of a zinc electrode immersed in a solution containing Zn2+ ions. The reduction reaction at the cathode involves the reduction of IO3- ions and H+ ions to form I2 and water:
2IO3- + 12H+ + 10e- ---> I2 + 6H2O
The oxidation reaction at the anode involves the oxidation of zinc atoms to form Zn2+ ions:
Zn ---> Zn2+ + 2e-
Since the oxidation half-reaction involves the loss of 2 electrons, we need to multiply it by 5 to balance the number of electrons transferred in the whole cell reaction:
5Zn ---> 5Zn2+ + 10e-
To write the net cell reaction, we can combine the reduction and oxidation half-reactions and make sure the electrons cancel out:
2IO3- + 12H+ + 5Zn ---> I2 + 6H2O + 5Zn2+