Question

For the cell, Cu | Cu²⁺ (0.13 M) || Ag⁺ (0.01 M) | Ag,

1. Calculate the reduction potential of each electrode if the standard reduction potential for copper and silver electrodes are 0.34 V and 0.80 V respectively.
2. Calculate the emf of the cell.
3. Write the cell reaction.
4. Is this cell reaction spontaneous? Why?

Answer 1

Given: Cu | Cu²⁺ (0.13 M) || Ag⁺ (0.01 M) | Ag

\[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = +0.34 V

\[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V

Temperature = 25°C (298 K)

Use the Nernst equation:

\[\ce{E = E{^{\circ}} - \frac{0.0591}{n} log \frac{1}{[ion]}}\]

a. Reduction potential of Ag⁺/Ag electrode (Right side):

\[\ce{E_{Ag^+/Ag} = 0.80 - \frac{0.0591}{1} log \frac{1}{0.01}}\] 

= 0.80 − 0.0591 × (−2)

= 0.80 + 0.118

= 0.9182 V

Reduction potential of Cu²⁺/Cu electrode (Left side):

\[\ce{E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} log \frac{1}{0.13}}\] 

= 0.34 − 0.02955 × (−log 0.13)

= 0.34 − 0.02955 × (−0.886)    ...(log 0.13 ≈ −0.886)

= 0.34 + 0.0262

= 0.3662 V

b. \[\ce{E{^{\circ}_{cell}} = E_{cathode} - E_{anode}}\]

= \[\ce{E_{Ag^+/Ag} - E_{Cu^{2+}/Cu}}\]

= 0.9182 − 0.3662

= 0.552 V

c. Cell reaction:

Since silver has a higher reduction potential, Ag⁺ is reduced, and Cu is oxidised.

\[\ce{Cu_{(s)} + 2Ag^+_{ (aq)} −> Cu^{2+}_{ (aq)} + 2Ag_{(s)}}\]

d. The cell potential is positive: (E_cell) = +0.552 V

This indicates the Gibbs free energy change is negative:

ΔG = −nFE_cell < 0

∴ The reaction is spontaneous.