Answers to Electrode Potentials 2

The cell diagram is:

Ag | AgCl | Cl2(g) | Pt

The cell is reversible in both solid and liquid phases.

RHE: ½Cl2(g) + e- ⇌ Cl-(s/l)

LHE: AgCl(s/l) + e- ⇌ Ag(s/l) + Cl-(s/l)

Formal Cell Reaction:

½Cl2(g) + Ag(s) ⇌ AgCl(s/l)

We know that G = H - TS and H = U +PV

So, G = U + PV - TS

Therefore, dG = dU + PdV + VdP - TdS - SdT

But, dU = dq + dw (where the bold derivatives remind us that these two quantities are non-state functions. [See Price, G, "Thermodynamics of Chemical Processes"; Oxford Chemistry Primer, OUP for a treatment of state functions.]

Under reversible conditions, dS = dqrev/T, and thus, considering only expansion work,

dU = TdS - PdV and so dG = VdP - SdT

Hence, (∂G/∂T)P = -S, or (∂ΔG/∂T)P = - ΔS

(where the ∂ refers to a partial derivative. See Sivia, ibid. for more about partial derivatives.)

Since ΔG = -FE for a one-electron process,

ΔS = F(∂E/∂T)P

Therefore, a graph of E against T will enable us to calculate ΔS.

The reader is recommended to read pp. 63-66 of the book, to see how to solve this particular problem. Readers looking for innovative and novel ways to solve it are invited to read on!

Harned Cell:

Pt | H2(g) (P= 1atm) | HCl(aq) | AgCl(s) | Ag(s)

Potential determining Equilibria

RHE: AgCl(s) + e- ⇌ Ag(s) + Cl-(aq)

LHE: H+(aq) + e- ⇌ ½H2(g)

Cell Reaction:

AgCl(s) + ½H2(g) ⇌ Ag(s) + H+(aq) + Cl-(aq)

Nernst Equation: Ecell = E0cell - {RT/F}ln{aH+ aCl- a Ag / aAgCl PH2 ½}

Notice that since hydrogen is a gas at the temperature of the experiment (298K), we use its partial pressure. ( Strictly, we should talk about the fugacity of hydrogen. This is similar to the concept of activity. However, we can consider hydrogen [to an excellent approximation] as being an ideal gas, with a fugacity coefficient equal to unity.)

Also, since silver and silver chloride are pure solids, their activities can be equated to unity.

Therefore, the Nernst Equation simplifies to: 

Ecell = E0cell - {RT/F}ln{aH+ aCl-}

Now, aH+ = f+mH+ = f+mHCl, and aCl- = f-mCl- = f-mHCl

But in the electrolyte solution, the H+ ions and Cl- ions co-exist, and hence, the activity coefficients of the individual ions have no physical meaning. We therefore use the mean ionic activity coefficient f±, which is defined on p.63, for each ionic species.

Therefore, since we are dealing with a 1:1- electrolyte, f±= (f+ f-)½

Hence, substituting these results into the Nernst Equation, and changing from natural logarthims to decadic logarthims gives:

Ecell = E0cell - 2.303{RT/F}log mHCl2 - 2.303{RT/F}log {f±2}

E0cell = E0Ag|AgCl - E0H+|H2 = E0Ag|AgCl  ; (by definition of E0H+|H2 = 0V)

Thus, Ecell = E0Ag|AgCl - 2.303{RT/F}log mHCl2 - 2.303{RT/F}log {f±2}

The Debye-Hückel Limiting Law (DHL) states that logf± = -A|z+||z-|I½, where z+, z- are the charges on the ions, A is a temperature and solvent dependent parameter (= 0.509 in water at 298K) and I is the ionic strength of the solution, as defined on p.45. In this case, I = m. [The committed reader will find a derivation of the DHL in chapter 2, Rieger, P H, "Electrochemistry"; Prentice-Hall, New Jersey (1987).]

Hence, we can calculate a value for log f± .

m / mol kg-1 I / mol kg-1 log f±

4 x 10-4 4 x 10-4 -0.01

9 x 10-4 9 x 10-4 -0.015

1.6 x 10-3 1.6 x 10-3 -0.02

2.5 x 10-3 2.5 x 10-3 -0.025

We can now determine a value for the Standard Electrode Potential for the silver/silver chloride couple.

Ecell = E0Ag|AgCl - 2.303(2){RT/F}ln m - 2.303(2){RT/F}log {f±}

or, Ecell + 4.606{RT/F}log {f±} = E0Ag|AgCl - 4.606{RT/F}log m

We now plot a graph of Ecell + 4.606{RT/F}log {f±} against log m. This gives an intercept of EAg|AgCl.

E / V I / mol kg-1 Ecell + 4.606{RT/F}log {f±} log m

0.62565 4 x 10-4 0.62447 -3.398

0.58460 9 x 10-4 0.58283 -3.046

0.55565 1.6 x 10-3 0.55329 -2.796

0.53312 2.5 x 10-3 0.53020 -2.602

Both cells A and B are Harned Cells

Pt | H2(g) | HCl(aq) | AgCl(s) | Ag(s)

Potential determining Equilibria:

RHE: AgCl(s) + e- ⇌ Ag(s) + Cl-(aq)

LHE: H+(aq) + e- ⇌ ½H2(g)

Cell Reaction:

AgCl(s) + ½H2(g) ⇌ Ag(s) + H+(aq) + Cl-(aq)

Nernst Equation: Ecell = E0cell - {RT/F}ln{aH+ aCl-},

for PH2 = 1 bar which approximates to 1 atm., and where E0cell = E0Ag|AgCl (by definition of E0H+|H2 = 0V).

Consider Cell A

aH+ = Kw/ aOH-

Hence, Ecell = E0Ag|AgCl - {RT/F}ln{Kw aCl-/ aOH-}

Now, aX = fX[X], where fX is the activity coefficient of X (also given the lower case Greek symbol gamma) and, [Cl-] = [OH-] = c

Also, since c < 10-2M, the Debye-Hückel Limiting Law (DHL) applies.

Since lgfi = -Azi2I½, and I, zi2 and A are all the same for both Cl- and OH-, fCl- = fOH- and so,

Ecell = E0Ag|AgCl - {RT/F}lnKw (1)

From the data given, we notice that Ecell A is not a function of concentration (as equation (1) predicts), and therefore,

1.0493 = E0Ag|AgCl - {RT/F}lnKw (2)

We next need a value for E0Ag|AgCl which (if we are to find Kwmust therefore be obtained from cell B.

Now we consider Cell B:

Ecell = E0Ag|AgCl - {RT/F}ln{aH+ aCl-}

aH+ = f+c and aCl- = f-(2c)

Using the DHL, we see that f+ = f- = f , since zi2, A(= 0.509 in water at 298K) and I are the same for both H+ and Cl-, as they co-exist in a solution of ionic strength I.

I = ½(c + 2c) = 3c/2 mol dm-3

Therefore, E = E0Ag|AgCl - {RT/F}ln{f2 2c2}

E + {RT/F}ln{2c2} = E0Ag|AgCl - {4.606RT/F}lg(f)

E + {RT/F}ln{2c2} = E0Ag|AgCl - 4.606(0.509)RT(3/2)½ c½/ F

Hence, a plot of E + {RT/F}ln{2c2} (on the ordinate axis) against c½ (on the abscissa) will have an ordinate intercept equal to the Standard Electrode Potential for Ag|AgCl.

E/V c/mol dm-3 c½/ mol½ dm-3/2 E + {RT/F}ln{2c2}/V

0.578 0.001 0.0316 0.2411

04969 0.005 0.0707 0.2426

0.4624 0.010 0.1000 0.2437

Cell diagram:

Pt | H2 (1 atm) | NH3 (c), NH4Cl (c) | AgCl | Ag

This is (again) a Harned Cell ie.

Pt | H2 | HCl | AgCl |Ag

Potential determining Equilibria:

RHE: AgCl(s) + e- ⇌ Ag(s) + Cl-(aq)

LHE: H+(aq) + e- ⇌ ½H2(g)

Cell Reaction:

AgCl(s) + ½H2(g) ⇌ Ag(s) + H+(aq) + Cl-(aq)

Nernst Equation: Ecell = E0cell - {RT/F}ln{aH+ aCl-},

for PH2 = 1 bar (which approximates to 1 atm., and where E0cell = E0Ag|AgCl (by definition of E0H+|H2 = 0V).

Consider the acid dissociation equilibrium for the ammonium ion:

NH4+(aq) ⇌ NH3(aq) + H+(aq)

The acid dissociation constant for this equilibrium is: 

Ka = aNH3 aH+ / aNH4+

We notice from the cell diagram that [NH4+] = [NH3] = c.

Thus, Ka = aH+ fNH3 / fNH4+. We can assume that fNH3 = 1. This is very reasonable, because ammonia is an uncharged species. Therefore, Ka = aH+ / fNH4+, which implies that:

aH+ = fNH4+ Ka

Thus, E = 0.2223 - {RT/F}ln{c Ka fNH4+ fCl-}

Now, c < 10-2M, and so the DHL applies. (lg fi = -Azi2 I½) Since zi2 and I are the same for both NH4+ and Cl-,

fNH4+ = fCl- =f± (since HCl is a 1:1 electrolyte) = f

Hence, E + {RT/F}ln(c) - 0.2223 = +4.606RT A I½ / F - {RT/F}ln{Ka}

We now plot a graph of E + {RT/F}ln(c) - 0.2223 against I½ (= c½). This should be a straight line with ordinate-intercept -{RT/F}ln{Ka}.

c / M E / V c / mol dm-3/2 E + {RT/F}ln(c) - 0.2223 / V

0.00857 0.8965 0.09257 0.55198

0.01580 0.8824 0.12570 0.55359

0.02120 0.8757 0.14560 0.55464

0.03030 0.8661 0.18166 0.55620

The reader is advised to re-read pp. 91-92 before considering this solution.

Cell diagram:

Cd(amalgam, 4.6% Cd) | CdCl2(aq, cM) | AgCl(s) | Ag(s)

Potential determining equilibria:

RHE: AgCl(s) + e- ⇌ Ag(s) + Cl-(aq)

LHE: ½Cd2+ + e- ⇌ ½Cd(amal)

Cell Reaction:

AgCl(s) + ½Cd(amal) ⇌ Ag(s) + ½Cd2+(aq) + Cl-(aq)

The Nernst Equation for this cell reaction is: E = E0cell - {RT/F}ln {aCl- aCd2+½}

(where the activity of cadmium in the amalgam is taken to be unity).

aCl- = f-(2c) and aCd2+ = f+(c). Since c < 10-2M, we can use the DHL. The ionic strength of the solution, I = ½[ 4c +2c] = 3c. Since this is the same for both Cd2+ and Cl-, and since the ratio

zCd2+2/zCl-2 = 4, we can deduce that lg f+/lg f- = 4. ie., f+ = f-4. We can use this result to determine E0cell. However, it is easier and more convenient to use the mean ionic activity coefficient, f±, which is defined on p.63.

f± 3 = f-2 f+ = f-6

Hence, f± = f-2

Hence, aCl- = f±(2c) = f±[Cl-] and aCd2+ = f±(c) = f±[Cd2+]

Therefore, 

E = E0cell - {RT/F}ln{f±3/2 (2)(c)3/2}

or,

E + {RT/F}ln{2c3/2} = E0cell - {3 RT/ 2F}ln{f±} (*)

or, using DHL, 

E + {RT/F}ln{2c3/2} = E0cell - {6.909 RT A 31/2 / 2F} c1/2

We plot a graph of E + {RT/F}ln{2c3/2} against c1/2, and extrapolate to find the ordinate intercept (ie. a value for E0cell). The cell emf when aCd2+ = aCl- = 1 corresponds to Ecell.

c / 10-3 mol dm-3 E / V c1/2 / mol1/2 dm-3/2 E + {RT/F}ln{2c3/2} / V

0.1087 0.9023 0.01041 0.56861

0.1269 0.8978 0.01126 0.57007

0.2144 0.8803 0.01464 0.57277

0.3659 0.8641 0.01913 0.57715

(i) The half-cell reaction for the anthraquinone monosulphonate redox-couple is

1/2 A(aq) + e- + H+ ⇌ 1/2 AH2 (aq)

which implies the following electrochemical cell

Pt|H2(g) (P=1 atm) | H+(aq) (a=1) ||A(aq)(aA), H+(aq) (aH+), AH2 (aq) (aAH2) |Pt

The corresponding Nernst equation is as follows noting that the hydrogen electrode above is standard

                                           E = E0A/AH2 + RT/F ln[(aA1/2 aH+)/( aAH21/2)]

Since anthraquinone and anthrahydroquinone are neutral molecules and uncharged their activities can, to a good degree of accuracy, by approximated by their concentrations. The Nernst equation is simplified to

                                           E = E0A/AH2 + RT/F ln[ ([A]1/2 aH+)/( [AH2]1/2) ]

 (ii) The Nernst equation from (i) can be written as

E = E0A/AH2 + RT/F ln[ ([A]1/2 aH+)/( [AH2]1/2) ] + RF/T ln(aH+)

Substituting log10N = lnN/ln(10) into the above gives

E = E0A/AH2 + RT/F ln[ ([AH2]1/2)/( [A]1/2) ] + ln(10)RF/T log10(aH+)

Noting that pH = -log10(aH+), the Nernst equation can be written in terms of pH

E = E0A/AH2 + RT/F ln[ ([AH2]1/2)/( [A]1/2) ] + ln(10)RF/T pH                  

Differentiating the above equation with respect to pH gives

                                  dE/d(pH) = -ln(10)RT/F

At 25°C, T = 298K.

                                  dE/d(pH) = -ln(10) x 298 x 8.31 / 96485.3 = -59mV per pH unit

(iii) The generalised redox process

B + ne- + mH+ ⇌ BHm

can be written as an one electron half-cell reaction

1/n B + e- + m/n H+ ⇌ 1/n BHm

The Nernst equation for the generalised redox process is therefore

E = E0B/BHm+ RT/F ln[ (aB1/n aH+m/n)/( aBHm1/n) ]

E = E0B/BHm+ RT/F ln[ (aB1/n)/( aBHm1/n) ] + RF/T ln(aH+m/n)

E = E0B/BHm+ RT/F ln[ (aB1/n)/( aBHm1/n) ] + mRF/nT ln(aH+)

E = E0B/BHm+ RT/F ln[ (aB1/n)/( aBHm1/n) ] + mRF/nT log10(aH+)

Substituting in pH = -log10(aH+) gives

E = E0B/BHm+ RT/F ln[ (aB1/n)/( aBHm1/n) ] + mRF/nT pH

Differentiating the above equation with respect to pH gives

dE/d(pH) = - mRT/nF ln(10)

(iv) Acid dissociation constant, KA, for the dissociation reaction HA⇌ H+ + A-  is given by

KA = aH+ aA- / aHA

(v) For A/AH2 system (e.g. anthraquinone/anthrahydroquinone), m = 2, n = 2

                            slope  = dE/d(pH) = 2RT/2nF ln(10) = -59mV per pH unit

                              pKa1 (AH-/AH2) = -log(2x10-8)=7.7; pKa2 (A2-/AH-) = -log(1.2x10-11)=10.9

Accordingly, the reduced species (anthrahydroquinone) exists predominantly as AH2 for pH less than 7.7, as AH- in the pH range 7.7 to 10.9 and as A2- for pH > 10.9. Applying the results derived above leads to the following E – pH behaviour using the value n = 2 throughout but m = 0, 1 or 2 as appropriate.

(vi, vii)