Question

Based on the reaction, Sn + Cu²⁺ ⇌ Sn²⁺ + Cu⁺ (in acidic medium)

Calculate the following values:

a.      E°cell

b.     ΔG°

c.      K

d.     E (when [Cu²⁺] = 0.0200 M, [Sn²⁺] = 0.0100 M, and [Cu⁺] = 0.0300 M)

 

Answer 1

Question:

Based on the reaction, Sn + Cu²⁺ ⇌ Sn²⁺ + Cu⁺ (in acidic medium)

Calculate the following values:

a.      E°cell

b.     ΔG°

c.      K

d.     E (when [Cu²⁺] = 0.0200 M, [Sn²⁺] = 0.0100 M, and [Cu⁺] = 0.0300 M)

Solution:

The balanced reaction is:

Sn + 2Cu²⁺ ⇌ Sn²⁺ + 2Cu⁺

The half reactions occurring on the electrodes are:

Sn - 2e^- $\rightarrow$ Sn²⁺

Cu²⁺ + e^- $\rightarrow$ Cu⁺

Thus, the first reaction is the oxidation ( so the electrode is the anode) and the second reaction is the reduction (so the electrode is cathode).

The standard cell potential E°cell is the difference of the standard reduction potential of the electrodes:

$E°_{cell} = E°_{red, cathode} - E°_{red, anode}$

The standard potentials for the half reactions shown above are:

Sn - 2e^- $\rightarrow$ Sn²⁺ $E°_{red, anode} = -0.137$ V

Cu²⁺ + e^- $\rightarrow$ Cu⁺ $E°_{red, cathode} = 0.340$ V

Therefore, the standard potential of the cell is:

$E°_{cell} = 0.340 - (-0.137) = 0.477$ V

The change in free energy $\Delta G$ is related to the standard potential as follows:

$∆G = -nFE°$

where $nF$ is the charge transferred during the reaction, 2·96485 J/V. Therefore:

$∆G = -2·96485·0.477 = -92$ kJ.

The relationship between the change in free energy and the equilibrium constant is:

$∆G=-RTlnK$

$K = \text{exp}(-\frac{∆G}{RT}) = \text{exp}(\frac{92047}{8.314·298}) = 1.4·10^{16}$

According to the Nernst equation, the electrochemical cell potential depends on the concentrations of the reactants and products:

$E = E° - \frac{RT}{nF}\text{log}\frac{[Sn^{2+}][Cu^+]^2}{[Cu^{2+}]^2}$

$E = 0.477 - \frac{8.314·298}{2·96485}\text{log}\frac{0.01·0.03^2}{0.02^2} = 0.498$ V.

Answer:

a. E°cell = 0.477 V

b.     ΔG° = -92 kJ

c.     K = 1.4·10¹⁶

d.     E (when [Cu²⁺] = 0.0200 M, [Sn²⁺] = 0.0100 M, and [Cu⁺] = 0.0300 M) = 0.498 V