Answer:
1: Anode: Zn ⇒ Zn²⁺ + 2e
Cathode: 2Ag ⁺ + 2e⁻ ⇒ 2Ag⁻
2: 1.56 V
3: -3.01 x 10⁵ J
4: Decrease
Step-by-step explanation:
From a reference table we can get the standard reduction potentials:
Ag ⁺ + e⁻ ⇒ Ag Eº = 0.80 V
Zn²⁺ + 2 e⁻ ⇒ Zn Eº = -0.76 V
The Ag will be reduced since it has the more positive standard reduction potential, and Zn will be oxidized.
The balanced half reactions are:
Zn ⇒ Zn²⁺ + 2e⁻ (anode where oxidation occurs) Eº = - (-0.76 V)= +0.76V
2Ag ⁺ + 2e⁻ ⇒ 2Ag Eº = 0.80 V (cathode /reduction) Eº = 0.80 V
Note the reversal of sign for the Zn half cel,l and that even though we multiplied the Ag half cell by 2 to balance, we dont do the same with the reduction potential since it is an intensive property.
The standard cell voltage will be given by:
Eºcell = Eº anode + Eº cathode = + 0.76 V + 0.80 V = 1.56 V
Please be careful with this formula since there are texts that write it as Eºcell= Eº cathode - Eºanode which is equivalent since this will reverse the sign we did previously.
ΔGº = - nFEº where
n= number of moles of electrons electrons exchanged ,
F= Faraday constant, 96500 Coulombs/ V mol e⁻,
Eº = the standard cell voltage
ΔGº = - 2 mol e⁻ x 96500 coulombs/ V mol e x 1.56 V = -3.01 x 10⁵ J
To answer part 4 we need to remember that what we are doing is effectively adding the chloride ions to the half cells, and that the solubilty of silvel chloride is extremely low. Therefore some of the Ag⁺ ions will be precipitated, and the cell voltage will decrease.