Question

Consider the following cell reaction at 298 K:

\[\ce{2Ag+ + Cd -> 2Ag + Cd^{2+}}\]

The standard reduction potentials (E°) for Ag⁺/Ag and Cd²⁺/Cd are 0.80 V and −0.40 V respectively.

1. Write the cell representation.
2. What will be the emf of the cell if concentration of Cd²⁺ is 0.1 M and that of Ag⁺ is 0.2 M?
3. Will the cell work spontaneously for the condition given in (b) above?

Answer 1

Given: \[\ce{2Ag+ + Cd -> 2Ag + Cd^{2+}}\]

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

\[\ce{E^{\circ}_{Cd^+/Cd}}\] = −0.40

a. Since Cd is oxidised (anode) and Ag⁺ is reduced (cathode), the cell is:

\[\ce{Cd_{(s)} | Cd^{2+}_{ (aq)} || Ag^+_{ (aq)} | Ag_{(s)}}\]

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

= 0.80 − (−0.40)

= 1.20 V

Balanced reaction:

\[\ce{Cd + 2Ag+ −> Cd^{2+} + 2Ag}\]

n = 2 

Cd²⁺ = 0.1 M

Ag⁺ = 0.2 M

Nernst equation:

\[\ce{E = E^{\circ}_{cell} - \frac{0.0591}{n} log \frac{[Cd^{2+}]}{[Ag^+]^2}}\]

= \[\ce{1.20 - \frac{0.0591}{2} log \frac{0.1}{(0.2)^2}}\]

= \[\ce{1.20 - 0.02955 log \frac{0.1}{0.04}}\]

= 1.20 − 0.02955 log (2.5)

= 1.20 − 0.02955 × 0.398    ...(log (2.5) ≈ 0.398)

= 1.20 − 0.01176

= 1.188 V

c. Yes, because the EMF is positive (E_cell = 1.188 V). This means the Gibbs free energy change (ΔG = −nFE_cell) is negative, so the reaction is spontaneous.