Question

4.The voltaic cell has with Pt/H+/H2 and Ag/AgC/Cl- half cells is a possible design for an electronic pH meter, in that the actual cell E depends on [H3O+].

(a) Write out (under each half cell) the electrode reactions, and give below the overall cell equation.
(b) Indicate with arrows the direction of motion of the ions and electrons as the cell reacts spontaneously.
(c) Mark the electrodes as + or – and cathode or anode.
(d) What is the standard cell potential, Eo?
Eo = _______________________
(e) Calculate the actual cell potential, E, if the unknown [H3O+] is 1.0 x 10-4 M.
E = _________________________
(f) If [H+] remains variable, then for this cell E = A + B.pH. What are the values of the Constants A and B?
A = ____________ , B = ______________

Answer 1

(a) Electrode reactions:

Pt/H+/H2: 2H+(aq) + 2e- -> H2(g) (reduction)

Ag/AgCl/Cl-: AgCl(s) + e- -> Ag(s) + Cl-(aq) (reduction)

Overall cell equation: 2AgCl(s) + H2(g) -> 2Ag(s) + 2HCl(aq)

(b) Direction of motion of ions and electrons:

In the Pt/H+/H2 half-cell, hydrogen ions (H+) move towards the platinum electrode and accept electrons to form hydrogen gas (H2). In the Ag/AgCl/Cl- half-cell, silver ions (Ag+) move towards the silver chloride (AgCl) electrode and accept electrons to form silver (Ag) metal while chloride ions (Cl-) move away from the electrode. Electrons move from the hydrogen electrode to the silver electrode through the external circuit.

(c) Electrode labeling:

The Pt/H+/H2 electrode is the cathode (-) and the Ag/AgCl/Cl- electrode is the anode (+).

(d) Standard cell potential (Eo):

The standard cell potential can be calculated using the standard reduction potentials for each half-cell:

Eo(cell) = Eo(reduction, Ag/AgCl/Cl-) - Eo(reduction, Pt/H+/H2)

Eo(reduction, Ag/AgCl/Cl-) = +0.222 V (from standard reduction potential tables)

Eo(reduction, Pt/H+/H2) = 0 V (by definition)

Eo(cell) = +0.222 V - 0 V = +0.222 V

(e) Actual cell potential (E):

E(cell) = Eo(cell) - (0.0592 V / n) * log[H3O+]

where n is the number of electrons transferred in the balanced equation (2 in this case)

E(cell) = +0.222 V - (0.0592 V / 2) * log(1.0 x 10^-4 M)

E(cell) = +0.222 V - (0.0296 V) = +0.1924 V

(f) Values of constants A and B:

E(cell) = A + B.pH

At pH 7 (neutral), E(cell) = Eo(cell) = +0.222 V

Therefore, A = +0.222 V and B = -0.0592 V/pH