Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap10

E10.2b The potential difference between the electrodes in a working electrochemical cell is called the cell potential.. The cell potential is not a constant and changes with time as the

Solutions to exercises

Discussion questions

E10.1(b) The Debye–H¨uckel theory is a theory of the activity coefficients of ions in solution It is the coulombic

(electrostatic) interaction of the ions in solution with each other and also the interaction of the ionswith the solvent that is responsible for the deviation of their activity coefficients from the ideal value

of 1 The electrostatic ion–ion interaction is the stronger of the two and is fundamentally responsiblefor the deviation Because of this interaction there is a build up of charge of opposite sign around anygiven ion in the overall electrically neutral solution The energy, and hence, the chemical potential

of any given ion is lowered as a result of the existence of this ionic atmosphere The lowering of thechemical potential below its ideal value is identified with a non-zero value ofRT ln γ± This non-zerovalue implies thatγ±will have a value different from unity which is its ideal value The role of thesolvent is more indirect The solvent determines the dielectric constant,, of the solution Looking

at the details of the theory as outlined in Justification 10.2 we see that  enters into a number of the

basic equations, in particular, Coulomb’s law, Poisson’s equation, and the equation for the Debyelength The larger the dielectric constant, the smaller (in magnitude) is lnγ±

E10.2(b) The potential difference between the electrodes in a working electrochemical cell is called the cell

potential The cell potential is not a constant and changes with time as the cell reaction proceeds.

Thus the cell potential is a potential difference measured under non-equilibrium conditions as electric

current is drawn from the cell Electromotive force is the zero-current cell potential and corresponds

to the potential difference of the cell when the cell (not the cell reaction) is at equilibrium

E10.3(b) The pH of an aqueous solution can in principle be measured with any electrode having an emf that is

sensitive to H+(aq) concentration (activity) In principle, the hydrogen gas electrode is the simplestand most fundamental A cell is constructed with the hydrogen electrode being the right-hand electrodeand any reference electrode with known potential as the left-hand electrode A common choice isthe saturated calomel electrode The pH can then be obtained from eqn 10.43 by measuring the emf(zero-current potential difference), E, of the cell The hydrogen gas electrode is not convenient to

use, so in practice glass electrodes are used because of ease of handling

Numerical exercises

E10.4(b) NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

NaCl, AgNO3and NaNO3are strong electrolytes; therefore the net ionic equation is

Ag+(aq) + Cl−(aq) → AgCl(s)

rH−−= f H−−(AgCl, s) − fH−−(Ag+, aq) − fH−−(Cl−, aq)

= (−127.07 kJ mol−1) − (105.58 kJ mol−1) − (−167.16 kJ mol−1)

KS=b(Pb b−−2+)×b(S b−2−−) b(Pb2 +) = b(S2 −) = S

181 pm

189 pm = 0.958

We havehydG−−(Cl−) = −379 kJ mol−1[Exercise 10.6a]

SohydG−−(NO−3) = (0.958) × (−379 kJ mol−1) = −363 kJ mol−1

Question Can you establish that the statement in the comment following the solution to Exercise

10.8a (in the Student’s Solutions Manual) holds for the solution of this exercise?

and(0.890 mol kg−1) × (0.500 kg) = 0.445 mol KNO3

So(0.445 mol) × (101.11 g mol−1) = 45.0 g KNO3 must be added

and(0.2967 mol kg−1) × (0.500 kg) = 0.1484 mol Ba(NO3)2

So(0.1484 mol) × (261.32 g mol−1) = 38.8 g Ba(NO3)2

E10.10(b) I (Al2(SO4)3) = 1

0.524 × 100 per cent = 47.1 per cent

E10.14(b) The extended Debye–H¨uckel law is log γ±= −A|z+z−|I1/2

1+ BI1/2

Solving forB

B = −

1

(b/b−  −)1/2 + 0.509

logγ±



Draw up the following table

rG−−= f G−−(PbI2, aq) − fG−−(PbI2, s)

fG−−(PbI2, aq) = rG−−+ f G−−(PbI2, s)



= +56.3 mV

E10.17(b) Identify electrodes using species with the desired oxidation states.

L: Cd(s) + 2OH−(aq) → Cd(OH)2(s) + 2e−

R: Ni(OH)3(s) + e− → Ni(OH)2 (s) + OH−(aq)

Cd(s)|Cd(OH)2(s)|OH−(aq)|Ni(OH)2(s)|Ni(OH)3(s)|Pt

E10.18(b) The cell notation specifies the right and left electrodes Note that for proper cancellation we must

equalize the number of electrons in half-reactions being combined

(a) R: Ag2CrO4(s) + 2e−→ 2Ag(s) + CrO2 −

Overall(R − L): Sn4 +(aq) + 2Fe2 +(aq) → Sn2 +(aq) + 2Fe3 +(aq) −0.62 V

Overall(R − L): Cu(s) + MnO2(s) + 4H+(aq) → Cu2 +(aq) + Mn2 +(aq)

Comment Those cells for whichE−−> 0 may operate as spontaneous galvanic cells under standard

conditions Those for whichE−− < 0 may operate as nonspontaneous electrolytic cells Recall that

E−−informs us of the spontaneity of a cell under standard conditions only For other conditions werequireE.

E10.19(b) The conditions (concentrations, etc.) under which these reactions occur are not given For the purposes

of this exercise we assume standard conditions The specification of the right and left electrodes isdetermined by the direction of the reaction as written As always, in combining half-reactions to form

an overall cell reaction we must write half-reactions with equal number of electrons to ensure propercancellation We first identify the half-reactions, and then set up the corresponding cell

(a) R: 2H2O(l) + 2e− → 2OH−(aq) + H2(g) −0.83 V

and the cell is

Na(s)|Na+(aq), OH−(aq)|H2(g)|Pt +1.88 V

or more simply

Na(s)|NaOH(aq)|H2(g)|Pt

(b) R: I2(s) + 2e−→ 2I−(aq) +0.54 V

L: 2H+(aq) + 2e−→ H2(g) 0and the cell is

Pt|H2(g)|H+(aq), I−(aq)|I2(s)|Pt +0.54 V

or more simplyPt|H2(g)|HI(aq)|I2(s)|Pt

Comment All of these cells have E−− > 0, corresponding to a spontaneous cell reaction under

standard conditions IfE−−had turned out to be negative, the spontaneous reaction would have beenthe reverse of the one given, with the right and left electrodes of the cell also reversed

E10.20(b) See the solutions for Exercise 10.18(b), where we have used E−− = ER−− − EL−−, with standard

electrode potentials from Table 10.7

E10.21(b) See the solutions for Exercise 10.19(b), where we have used E−− = ER−− − EL−−, with standard

electrode potentials from Table 10.7

E10.22(b) In each case find E−− = ER−−− EL−−from the data in Table 10.7, then use

E10.23(b) (a) A new half-cell may be obtained by the process (3) = (1) − (2), that is

(3) 2H2O(l) + Ag(s) + e−→ H2(g) + 2OH−(aq) + Ag+(aq)

But,E3−−= E1−−− E2−−, for the reason that the reduction potentials are intensive, as opposed toextensive, quantities Only extensive quantities are additive However, therG−−values of thehalf-reactions are extensive properties, and thus

(b) The complete cell reactions is obtained in the usual manner We take(2) × (2) − (1) to obtain

2Ag+(aq) + H2(g) + 2OH−(aq) → 2Ag(s) + 2H2O(l)

(b) rG = −νFE[10.32] = −(2)×(9.6485×104C mol−1)×(0.4658 V) = −89.89 kJ mol−1

(c) logγ± = −|z+z−|AI1/2[19]= −(0.509) × (0.010)1/2[I = b for HCl(aq)] = −0.0509

E10.25(b) R: Fe2+(aq) + 2e−→ Fe(s)

L: 2Ag+(aq) + 2e− → 2Ag(s)

R− L: 2Ag(s) + Fe2 +(aq) → 2Ag+(aq) + Fe(s)

Therefore,rG−−(308 K) ≈ (239) + (10 K) × (−0.206 K−1) kJ mol−1≈ +237 kJ mol−1

E10.26(b) In each case ln K = νFE−−

L: 3AgCl(s) + 3e−→ 3Ag(s) + 3Cl−(aq) EL−− = 0.22 V

R− L: Co3 +(aq) + 3Cl−(aq) + 3Ag(s) → 3AgCl(s) + Co(s)

E−−= ER−−− EL−− = (0.42 V) − (0.22 V) = +0.20 V

E10.28(b) First assume all activity coefficients are 1 and calculate KS◦, the ideal solubility product constant

(1) AgI(s) +(aq) + I−(aq)

S(AgI) = b(Ag+) = b(I−) because all stoichiometric coefficients are 1.

ThusKS◦=b(Ag+)b(I−)

b− 2  − = S2

b− 2  − = (1.2 × 10−8)2= 1.44 × 10−16

(2) Bi2S3(s) 3 +(aq) + 3S2 −(aq)

b(Bi3 +) = 2S(Bi2S3) b(S2 −) = 3S(Bi2S3)

Neglect of activity coefficients is not significant for AgI and Bi2S3

E10.29(b) The Nernst equation applies to half-reactions as well as whole reactions; thus for

8H++ MnO−4(aq) + 5e−→ Mn2 +(aq) + 4H2O

E = E−−−RT

5F ln

a(Mn2 +) a(MnO−4)a(H+)8

E10.30(b) R: 2AgI(s) + 2e−→ 2Ag(s) + 2I−(aq) −0.15 V

R: AgI(s) + e−→ Ag(s) + I−(aq)

Overall(R − L): AgI(s) → Ag+(aq) + I−(aq)

Since the cell reaction is a solubility equilibrium, for a saturated solution there is no further tendency

It is convenient to give the solution for (b) first.

(b) KS= K−1= e−224≈ 10−98 , since the cell reaction is the reverse of the solubility equilibrium

Solutions to numerical problems

P10.1 We require two half-cell reactions, which upon subtracting one (left) from the other (right), yields the

given overall reaction (Section 10.4) The half-reaction at the right electrode corresponds to reduction,that at the left electrode to oxidation, though all half-reactions are listed in Table 10.7 as reductionreactions

Hence, a suitable cell would be

Pb(s)|PbSO4(s)|H2SO4(aq)|Hg2SO4(s)|Hg(l)

or, alternatively,

Pb(s)|PbSO4(s)|H2SO4(aq)H2SO4(aq)Hg2SO4(s)|Hg(l)

For the cell in which the only sources of electrolyte are the slightly soluble salts, PbSO4and Hg2SO4,the cell would be

Pb(s)|PbSO4(s)|PbSO4(aq)Hg2SO4(aq)|Hg2SO4(s)|Hg(l)

The potential of this cell is given by the Nernst equation [10.34]

2.303 =

E − E−−

2.303RT /F + 0.05114

E−− = ER−−− EL−−= E−−(AgCl, Ag) − E−−(H+/H2) = +0.22 V − 0 [Table 10.7]

We then draw up the following table with the more precise value forE−−= +0.2223 V [Problem 10.8]

See the original reference for a careful analysis of the precise data.

P10.7 The cells described in the problem are back-to-back pairs of cells each of the type

a(M+)a(X−) =

γ+b+

Forb = 0.09141, the extended Debye–H¨uckel law gives

We then draw up the following table

E/V −0.0220 0.0000 0.0263 0.0379 0.1156 0.1336

A more precise procedure is described in the original references for the temperature dependence of

E−−(Ag, AgCl, Cl−), see Problem 10.10.

P10.10 The method of the solution is first to determinerG−−,rH−−, andrS−−for the cell reaction

1

2H2(g) + AgCl(s) → Ag(s) + HCl(aq)

and then, from the values of these quantities and the known values offG−−,fH−−, andS−− forall the species other than Cl−(aq), to calculate fG−−,fH−−, andS−−for Cl−(aq).

∂E−−

∂T



p = (−6.4566 × 10−4V/◦C) × (◦C/K) = −6.4566 × 10−4V K−1

Hence, from equation(a)

rS−− = (−96.485 kC mol−1) × (6.4566 × 10−4V K−1) = −62.30 J K−1mol−1

andrH−− = r G−−+ T r S−−

= −(21.46 kJ mol−1) + (298.15 K) × (−62.30 J K−1mol−1) = −40.03 kJ mol−1

For the cell reaction

R = 0.999 997 01 (an extremely good fit)

, C= Pa m3

V or

m3

C = VPa

Pa×1.01325 × 105Pa

atm = 2.80 × 10−6V atm−1

= 2.80 × 10−3mV atm−1This compares closely to the result from the potential measurements

(c) A fit to a second-order polynomial of the form

E = a + bp + cp2yields

R = 0.999 997 11

This regression coefficient is only marginally better than that for the linear fit, but the uncertainty

in the quadratic term is> 200 per cent.

= b = 2.835 × 10−3mV atm−1to

∂E

∂p

max

(d) We can obtain an order of magnitude value for the isothermal compressibility from the value ofc.

Comment It is evident from these calculations that the effect of pressure on the potentials of

cells involving only liquids and solids is not important; for this reaction the change is only

∼ 3×10−6V atm−1 The effective isothermal compressibility of the cell is of the order of magnitudetypical of solids rather than liquids; other than that, little significance can be attached to the calculatednumerical value

P10.15 The equilibrium is

K = a(H2O)4a(V4O12−4)

a(H2VO4 −)4 ≈ γ (V4O12−4)b(V4O12−4)

γ (H2VO4 −)4b(H2VO4 −)4Letx be b(H2VO4 −); then b(V4O12 −4) = (0.010 − x)/4 Then the equilibrium equation can be

P10.18 The reduction reaction is

Sb2O3(s) + 3H2O(l) + 6e− → 2Sb(s) + 6OH−(aq) Q = a(OH−)6 ν = 6

F (pOHf− pOHi) = (59.17 mV) × (pOHf− pOHi)

pOHf= − log(0.050γ±) = − log 0.050 − log γ±= − log 0.050 + A(0.050) = 1.415

pOHi= − log(0.010γ±) = − log 0.010 − log γ±= − log 0.010 + A(0.010) = 2.051

Hence,E = (59.17 mV) × (1.415 − 2.051) = −37.6 mV

P10.19 We need to obtainrH−−for the reaction

1

2H2(g) + Uup+(aq) → Uup(s) + H+(aq)

We draw up the thermodynamic cycle shown in Fig 10.2

Data are obtained from Table 13.4, 14.3, 2.6, and 2.6b The conversion factor between eV and

which corresponds to +111 kJ mol−1

The electrode potential is therefore −r G−−

νF , withν = 1, or −1.15 V

Solutions to theoretical problems

P10.21 MX(s) +(aq) + X−(aq), Ks≈ b(M+)b(X−)



b−  −

Therefore, sinceγ2

±= e−4.606AC1/2 S≈ Kse−4.606AC

1/2

C

P10.25 The half-reactions involved are:

R: cytox+ e−→ cytred Ecyt−−

L: Dox+ e− → Dred ED−−

The overall cell reaction is:

R− L = cytox+ Dred red+ Dox E− −  − = Ecyt−−− ED−−

(a) The Nernst equation for the cell reaction is

E = E − RT F ln[cytred][Dox]



= ln

[cyt]ox[cyt]red

+ F

against ln

[cyt]ox[cyt]red

(b) Draw up the following table:

ln

[Dox]eq

against ln

[cytox]eq[cytred]eq



is shown in Fig 10.3 The intercept is

−1.2124 Hence

E−  − cyt = RT

F × (−1.2124) + 0.237 V

= 0.0257 V × (−1.2124) + 0.237 V

= +0.206 V

–5 –6 –5 –4 –3 –2 –1 0

P10.27 (a) molalityH2SO4 = b(d) = a(d − d25 ) + c(d − d25)2

whered is density in g cm−3at 25◦C,a = 14.523 mol kg−1(g cm−3)−1,

c = 25.031 mol kg−1(g cm−3)−2, andd25= 0.99707 g cm−3.For 1 kg solvent(mH 2 O= 1 kg):

8 6 4 2 0

Sulfuric Acid Solutions

Sulfuric Acid Solutions

Mass Percentage Sulfuric Acid

(b) cell: Pb(s) | PbSO4(s) | H2SO4(aq) | PbO2(s) | PbSO4(s) | Pb(s)

cathode: PbO2(s) + 3H+(aq) + HSO−4(aq) + 2e−→ PbSO4(s) + 2H2O(l)

Ecathode−− = 1.6913 V

anode: PbSO4(s) + H+(aq) + 2e−→ Pb(s) + HSO−4

Eanode−− = −0.3588 V

net: PbO2(s) + Pb(s) + 2H+(aq) + 2HSO−4(aq) → 2PbSO4(s) + 2H2O(l)

E−− = Ecathode−− − Eanode◦ = 2.0501V (eqn 10.38)

rG−− = −νFE−− = −(2)(9.64853 × 104C mol−1)(2.0501 V)

= −3.956 × 105C V mol−1 = −3.956 × 105J mol−1 = −395.6 kJ mol−1

fH−−values of Table 2.6 and the CRC Handbook of Chemistry and Physics are used in the

pH (eqn 9.29)

E = E−−− (0.05916V) νH

For the PbO2| PbSO4couple,PbO2(s) + 4H++ SO2 −