For the pair of species given:

Question

a. Lithium (Eº = -3.05 V) and silver (Eº= 0.80 V) 

b. Cadmium (Eº= -0.40 V) and copper (Eº= +0.34 V)

1) Write balance half-reactions and the overall spontaneous reaction.

2) Diagram the cell. Be as specific as possible including labeling electrodes with their charges and names (anode and cathode), showing the direction of electron flow in the circuit and showing the direction of cation and anion flow in the salt bridge.

3) Calculate the cell potential for the overall reaction.

4) Calculate the Gibbs free energy for the overall reaction.

5) Write the standard cell notation for the overall reaction.

6) Calculate K_eq for the spontaneous reaction.

Answer 1

Li⁺ + e- ==> Li(s) Eº = -3.05 V

Ag⁺ + e- ==> Ag(s) Eº = 0.80 V

(1). Overall reaction: Ag⁺ + Li(s) ==> Ag(s) + Li⁺

(2). Can't draw on this platform but anode will be Li and cathode will be Ag. Electron flow will be from anode to cathode.

(3). Cell potential = Eºcell = 0.80 V - (-3.05 V) = 3.85 V

(4). Gibbs free energy = ∆G = -nFE = - (1)(96500)(3.85) = joules

(5). Li(s) | Li⁺(aq) || Ag⁺(aq) | Ag(s)

(6). ∆Gº = -RT ln K and solve for K