Electrochemistry

Question

Consider the galvanic cell:

Pt(s) I H2(g, 0.100 bar) I H+(aq, pH=2.54) II Cl-(aq, 0.200 M) I Hg2Cl2(s) I Hg(l) I Pt(s)

(a) Which half-cell is the anode and which is the cathode? Find their standard electrode

potentials from Appendix C.

(b) What is the standard cell potential?

(c) What is the net cell reaction?

(d) What is the cell voltage?

(e) In what direction does the reaction go?

(f) What is the equilibrium constant for the reaction?

Answer 1

By convention, cell diagram notation follows this format:

anode(s) l anode (aq, M) ll cathode (aq, M) l cathode(s) where M is the concentration of the solute in molarity, and a vertical line l separates different phases. Pt is used as an inert electrode when the species being oxidized or reduced are not solids, so you can't clip a wire to them directly.

Using this logic, we can assume H2 / H+ as the anode and HgCl2 / Hg as the cathode. But to double check, we can look at the standard reduction potentials. In a galvanic cell, the species with the higher reduction potential should be reduced and at the cathode. According to https://ch302.cm.utexas.edu/echem/echem-cells/selector.php?name=std-red-potentials

Hg2Cl2 + 2e- --> 2Hg + 2Cl- +0.27 V

2 H+ + 2e- --> H2 0.00 V as expected for a standard hydrogen electrode

This confirms that mercury is our cathode and hydrogen is our anode (a).

b) the standard cell potential may be found using the equation E∘cell = E cat - E an

where E∘cell is the standard cell potential, E cat is the standard reduction potential of the cathode, and E an is the standard reduction potential of the anode. E∘cell = 0.27 V - 0.00 V = 0.27 V

c) The net cell reaction is the sum of both half cell reactions.

At the cathode, we have Hg2Cl2(s) + 2e- --> 2Hg(l) + 2Cl-(aq)

At the anode, we are losing electrons (oxidation) so we will flip the reaction to reflect this: H2 --> 2 H+ + 2e-

Putting them together, the electrons cancel, and we get Hg2Cl2(s) + H2(g) --> 2Hg(l) + 2H+(aq) + 2Cl-(aq)

d)The cell voltage is given by the equation E cell = E∘cell - RT/nF lnQ

where E cell is the actual cell voltage, E∘cell is once again the standard cell potential, R is the gas constant 8.314 J/molK, T is the temperature in Kelvin, n is the moles of electrons transferred per round of reaction, F is Faraday's constant 96485 C/mol, and Q is the reaction quotient. We will assume room temperature since it is not stated.

E cell = 0.27 V - (8.314 J/molK) (298K) / ((2 mol) (96485 C/mol) ln Q

Note that the expression for Q is the same as Keq, ie the concentration/partial pressures of the products divided by the concentration/partial pressures of the reactants, raised to their coefficient powers. The only difference is we are not at equilibrium. Remember that only gases and solutes affect Q and Keq. From our net reaction, Q = [H+]^2 [Cl-]^2)/ P H2. It looks a little strange, because there is a mixture of both gases and solutions, but the overall expression should have both concentration brackets and partial pressures. Now,

E cell = 0.27 V - (8.314 J/molK) (298K) / ((2 mol) (96485 C/mol) ln ([H+]^2 [Cl-]^2)/ P H2)

[H+] = 10^-pH = 10^-2.54 = 0.002884

P H2 should be in atm so 0.100 bar x 0.987 atm/bar = 0.0987 atm

E cell = 0.27 V - (8.314 J/molK) (298K) / ((2 mol) (96485 C/mol) ln (0.002884^2 x 0.200^2/ 0.0987)

E cell = 0.27 V - (8.314 J/molK) (298K) / ((2 mol) (96485 C/mol) ln (3.37 x 10^-6)

E cell = 0.27 V - (8.314 J/molK) (298K) / ((2 mol) (96485 C/mol) (-12.6)

E cell = 0.27 V - (0.0128 V) (-12.6) (note that 1 V = 1 J/C)

E cell = 0.27 V + 0.161 V

E cell = 0.43 V

This makes sense, as our Q was quite a bit smaller than 1, meaning equilibrium is also pushing the reaction forward as well as the standard potential, raising the cell potential and making the reaction more likely than under standard conditions.

e) The reaction goes forward as written, because this is a galvanic cell, which means it will be a spontaneous downhill process. Whatever has a higher reduction potential will be reduced. The opposite would occur in an electrolytic cell, where energy is used to drive a nonspontaneous reaction.

f) as stated above, the equilibrium constant Keq has the same expression as Q, or [H+]^2 [Cl-]^2)/ P H2. The value for Keq at a certain temperature can be determined experimentally, looked up in a reference table (such as Ka or Kb at 25 C) or determined from ΔG (Gibbs free energy). Assuming it is Keq that is desired in this problem and not Q,

ΔG∘=−nFE∘cell and

ΔG°=−RTlnKeq so

nFE∘cell = RTlnKeq or ln Keq = nF/RT x E∘cell

ln Keq = nF/RT x E∘cell

ln Keq = 1/0.0128 V x 0.27 V

ln Keq = 21.0

Keq = e^21 = 1.36 x 10^9