Question

Consider the reaction:

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

The reduction potentials of Ag⁺/Ag and Cd²⁺/Cd are +0.80 volt and −0.40 volt respectively.

1. Give the cell representation.
2. What is the standard cell emf, E°?
3. What will be the emf of the cell if concentration of Cd²⁺ is 0.1 M and Ag⁺ is 0.2 M?
4. Will the cell work spontaneously for the condition given in (c) 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

c. 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

d. 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.