Write the equations describing the electrode reactions and the net cell reaction for this electrochemical cell containing tin and silver:
Sn(s)|Sn2 (aq)||Ag (aq)|Ag(s)
what is the equation for cathode, anode, and net cell reaction
To determine the equations for the cathode, anode, and net cell reaction in an electrochemical cell, we need to first identify the half-reactions at each electrode.
In this case, the electrode on the left side of the double vertical line (||) is the anode, and the electrode on the right side is the cathode.
The half-reaction at the anode will involve the oxidation of the species present, while the half-reaction at the cathode will involve the reduction of the species present.
1. Anode half-reaction: Sn(s) → Sn2+(aq) + 2e^-
The solid tin (Sn) is oxidized, losing two electrons to form Sn2+ ions in solution.
2. Cathode half-reaction: Ag+(aq) + e^- → Ag(s)
The Ag+ ions in solution are reduced, gaining one electron to form solid silver (Ag).
Next, we need to balance the half-reactions to ensure the number of electrons gained or lost is the same on both sides.
Multiplying the anode half-reaction by 2 and the cathode half-reaction by 1 will balance the number of electrons:
Anode: 2Sn(s) → 2Sn2+(aq) + 4e^-
Cathode: 2Ag+(aq) + 2e^- → 2Ag(s)
Finally, we can combine these half-reactions to obtain the overall or net cell reaction by canceling out the electrons:
Net Cell Reaction: 2Sn(s) + 2Ag+(aq) → 2Sn2+(aq) + 2Ag(s)
Therefore, the electrode reactions and net cell reaction for this electrochemical cell containing tin and silver are:
Anode: Sn(s) → Sn2+(aq) + 2e^-
Cathode: Ag+(aq) + e^- → Ag(s)
Net Cell Reaction: 2Sn(s) + 2Ag+(aq) → 2Sn2+(aq) + 2Ag(s)