Question

Study the diagram given below that represents Cu-Ag electrochemical cell and answer the questions that follow.

Given \[\ce{E^0_{(Cu^{2+}/Cu)}}\] = 0.337V; \[\ce{E^0_{(Ag^+/Ag)}}\] = 0.799V

1. Write the cell reaction for the above cell.
2. Calculate the standard emf of the cell.
3. If the concentration of [Cu²⁺] is 0.1 M and E_cell is 0.422 V, at 25°C, calculate the concentration of [Ag⁺].
4. Calculate ΔG for the cell.

Answer 1

1. Anode: \[\ce{Cu -> Cu^{2+} + 2e-}\]

Cathode: \[\ce{2Ag+ + 2e- -> 2Ag }\]

Cell reaction: \[\ce{Cu + 2Ag+ -> Cu^{2+} + 2Ag}\]

2. Standard emf \[\ce{(E^\circ_{cell}) = E^\circ_{cathode} - E^\circ_{anode}}\]

= 0.799 − 0.337

= 0.462 V

3. `E_("cell") = E_("cell") ^\circ - (0.0591)/n log  (Cu^(2+))/([Ag^+]^2)`

`0.422 = 0.462 - (0.0591)/2 log  ([0.1])/([Ag^+]^2)`

`log  ([0.1])/([Ag^+]^2) = ((0.462 - 0.422))/0.0591 xx 2`

= 1.35

log[0.1] − 2log[Ag⁺] = 1.35

−1 − 2log [Ag⁺] = 1.35

2log [Ag⁺] = −1 − 1.35 = − 2.35

log [Ag⁺] = −1.125

[Ag⁺] = 7.5 × 10⁻² M

4. ΔG = −nFE_cel

= − 2 × 96500 × 0.422

= 81446 J mol⁻¹

= 81.446 kJ mol⁻¹