Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red,cat - E°red,an
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
[tex]E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V[/tex]
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
