Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics. Fall Semester Homework Problem Set Number 12 Solutions

Transcription

1 Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 017 Homework Problem Set Number 1 Solutions 1. (Based on McQuarrie and Simon, 13-1.) Write balanced half-cell chemical reactions, and balanced complete chemical reactions for the below electrochemical cells. Also, indicate the value of n that would be used to relate ΔG o r = nfe o for the balanced complete reaction. a. Pb(s) PbI (s) HI(aq) H (g) Pt(s) Anode: Pb(s) + I (aq) à PbI (s) + e Cathode: H + (aq) + e à H (g) Balanced: Pb(s) + H + (aq) +I (aq) à PbI (s) + H (g) Value of n: b. Cu(s) Cu(ClO 4 ) (aq) AgClO 4 (aq) Ag(s) Anode: Cu(s) à Cu + (aq) + e Cathode: Ag + (aq) + e à Ag(s) Balanced: Cu(s) + Ag + (aq) à Cu + (aq) + Ag(s) Value of n: c. In(s) In(NO 3 ) 3 (aq) CdCl (aq) Cd(s) Anode: In(s) à In 3+ (aq) + 3e Cathode: Cd + (aq) + e à Cd(s) Balanced: In(s) + 3Cd + (aq) à In 3+ (aq) + 3Cd(s) Value of n: 6 d. Sn(s) SnCl (aq) AgNO 3 (aq) Ag(s)

2 Anode: Sn(s) à Sn + (aq) + e Cathode: Ag + (aq) + e à Ag(s) Balanced: Sn(s) + Ag + (aq) à Sn + (aq) + Ag(s) Value of n:. (Based on McQuarrie and Simon, 13-4.) Given the electrochemical cell reaction H (g) + PbSO 4 (s) D H + (aq) + SO 4 (aq) + Pb(s), predict the effect of the following changes on the observed cell voltage (increase, decrease, or no effect): a. increase in the pressure of H Increased voltage (increased chemical potential of a reactant) b. increase in the size of the lead electrode No effect (solid lead activity is 1 irrespective of size of electrode) c. decrease in the ph of the cell electrolyte Decreased voltage (increased activity of a product; remember that a lower ph is a higher H + activity) d. dissolution of Na SO 4 in the cell electrolyte Decreased voltage (increased activity of a product; common SO 4 ion effect) e. decrease in the amount of PbSO 4 No effect (solid lead sulfate activity is 1 irrespective of size of electrode) f. dissolution of a small amount of NaOH in the cell electrolyte

3 Increased voltage (decreased activity of a product; addition of hydroxide will lead to higher ph which is a lower H + activity) 3. (Based on McQuarrie and Simon, 13-3.) What is the 98 K EMF for the electrochemical cell Pt(s) Sn + (aq; a = 0.00), Sn 4+ (aq; a = 0.400) Fe 3+ (aq; a = 0.300) Fe(s)? given the standard reduction potentials E o = 0.15 and V for the half cells Sn 4+ (aq) + e à Sn + (aq) and Fe 3+ (aq) + 3e à Fe(s), respectively. Given the provided standard reduction potentials, the overall standard cell potential is E o = E o ox + E o red = ( 0.15) + ( 0.036) = V [Note that the standard oxidation potential for the tin anode is the negative of that reaction s standard reduction potential.] A balanced chemical equation is 3Sn + (aq) + Fe 3+ (aq) à 3Sn 4+ (aq) + Fe(s) for which n = 6. To compute the cell potential for the indicated activities, we use the Nernst equation: E = E o RT a nf ln Sn a Sn + a Fe a Fe 3+ ( J mol 1 K 1 ) 98 K = V 6 ( )( 96,485 C mol 1 ) ( ) ln( 88.89) = V V = 0.05 V ( ) ( ) 3 ( 1) ( 0.) 3 ( 0.3) ln 0.4 Note that the negative cell potential implies that this cell will actually run in reverse; that is, solid iron will be oxidized and dissolve in order to reduce aqueous tin(iv) ions, even though the cell is written with the Pt-catalyzed Sn electrode at left in the anode position.

4 4. (Based on McQuarrie and Simon, ) What is the 98 K solubility product (K sp ) of PbSO 4 given the standard reduction potentials E o = and 0.16 V for the half cells PbSO 4 (s) + e à Pb(s) + SO 4 and Pb + (aq) + e à Pb(s), respectively. Given the provided standard reduction potentials, the overall standard cell potential for PbSO 4 (s) à Pb + (aq) + SO 4 (aq) is E o = E o ox + E o red = (0.16) + ( ) = 0.33 V [Note that the standard oxidation potential for the lead anode is the negative of that reaction s standard reduction potential.] For the above balanced chemical equation n =. At equilibrium, the cell potential will be zero and the Nernst equation: E = E o RT a a nf ln Pb + SO 4 a PbSO4 transforms to 0 = E o RT F ln K sp K sp = e FEo / RT plugging in appropriate constants (see last problem for values) provides the solution K sp = 1.39 x (Based on McQuarrie and Simon, ) What is the 98 K aciddissociation constant (K AH ) of HClO given the standard reduction potentials E o = 1.49 and 0.90 V for the half cells HClO(aq) + H + (aq) + e à Cl (aq) + H O(l) and ClO (aq) + H O(l) + e à Cl (aq) + HO (aq), respectively. You will need to use the water dissociation constant K w = to solve this problem.

5 We begin by constructing a balanced chemical equation for the two half reactions, e.g., HClO(aq) + H + (aq) + HO (aq) à ClO (aq) + H O(l). Given the provided standard reduction potentials, the overall standard cell potential is E o = E o ox + E o red = ( 0.90) + (1.49) = 0.59 V [Note that the standard oxidation potential for the chloride/hypochlorite anode is the negative of that reaction s standard reduction potential.] For the above balanced chemical equation n =. At equilibrium, the cell potential will be zero and the Nernst equation: E = E o RT nf ln a a ClO H O a HClO a H + a HO transforms to 0 = 0.59 V ( V) ln K# where K is defined by the ratio of activity coefficients in the Nernst equation above. The value of K is determined from the above equation as 8.8 x But we want to find the value for K AH = a ClO a H + a HClO which we can do by noting that

6 a ClO a H + a HClO = a a ClO H O a HClO a H + a HO a H + aho a H O = K$ K w ( )( 1 ) = = where we ve used the definition of K w for the water dissociation reaction H O(l) à H + (aq) + HO (aq). So, K AH is 8.8 x 10 9, i.e., hypochlorous acid (a constituent of chlorine bleach) is a relatively weak acid.