Answer:
E°cell = 0.94 V
Ecell = 1.00 V
ΔG = -1.9 × 10⁵ J
ΔG° = -1.8 × 10⁵ J
Explanation:
Let's consider this electrochemical cell:
Sn(s)|Sn²⁺(aq,0.0155M)||Ag⁺(aq, 3.50M)|Ag(s)
The corresponding half-reactions are:
Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻ E°red = -0.14 V
Reduction (cathode): 2 Ag⁺(aq) + 2 e⁻ → 2 Ag(s) E°red = 0.80 V
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an = 0.80 V - (-0.14 V) = 0.94 V
We can find the cell potential using the Nernst equation.
Ecell = E°cell - (0.05916/n) . log Q
Ecell = 0.94 V - (0.05916/2) . log ([Sn²⁺]/[Ag⁺]²)
Ecell = 1.00 V
We can find ΔG and ΔG° using the following expressions.
ΔG = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 × 10⁵ J
ΔG° = -n.F.E°cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 × 10⁵ J
