University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

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1 University Chemistry Quiz /04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell, what wuld be the standard cell ptential? Sl. (a) We can calculate G frm standard free energies f frmatin. G 2 G (N ) 6 G (H O) [4 G (NH ) 3 G (O )] f 2 f 2 f 3 f 2 G = 0 + (6)(237.2 kj ml 1 ) [(4)(16.6 kj ml 1 ) + 0] (b) The half-reactins are: G = kj ml 1 4NH3(g) 2N2(g) + 12H + (aq) + 12e O2(g) + 12H(aq) + 12e 6H2O(l) The verall reactin is a 12-electrn prcess. We can calculate the standard cell emf frm the standard free energy change, G. G nf cell cell G nf kj 1 ml 1000 J 1 kj (12)(96500 J V 1 ml 1 ) 1.17 V 2. (10%) Oxalic acid (H2C2O4) is present in many plants and vegetables. (a) Balance the fllwing equatin in acid slutin: MnO 4 + C 2 O 4 2 Mn 2+ + CO 2 (b) If a 1.00 g sample f H2C2O4 requires 24.0 ml f M KMnO4 slutin Sl. (a) t reach the equivalence pint, what is the percent by mass f H2C2O4 in the sample? The halfreactins are: (i) MnO4 (aq) + 8H (aq) + 5e Mn 2+ (aq) + 4H2O(l)

2 (ii) C2O4 2 (aq) 2CO2(g) + 2e We cmbine the half-reactins t cancel electrns, that is, [2 equatin (ii)] 2MnO4 (aq) + 16H (aq) + 5C2O4 2 (aq) 2Mn 2+ (aq) + 10CO2(g) + 8H2O(l) (b) We can calculate the mles f KMnO4 frm the mlarity and vlume f slutin ml KMnO 24.0 ml KMnO ml KMnO 1000 ml sln We can calculate the mass f xalic acid frm the stichimetry f the balanced equatin. The mle rati between xalate in and permanganate in is 5: ml H2C2O g H2C2O ( ml KMnO 4 4) g H2C2O4 2 ml KMnO4 1 ml H2C2O4 Finally, the percent by mass f xalic acid in the sample is: g % xalic acid 100% 5.40% 1.00 g 3. (5%) One f the half-reactins fr the electrlysis f water is 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e If L f O2 is cllected at 25 C and 755 mmhg, hw many mles f electrns had t pass thrugh the slutin? Sl. Find the amunt f xygen using the ideal gas equatin n PV RT 1 atm 755 mmhg 760 mmhg (0.076 L) ( Latm K 1 ml 1 )(298 K) ml O 2 Since the half-reactin shws that ne mle f xygen requires fur faradays f

3 electric charge, we write 3 4 F ( ml O 2) F 1 ml O2 4. (5%) Predict what will happen if mlecular brmine (Br2) is added t a slutin cntaining NaCl and NaI at 25 C. Assume all the species are in their standard states. Sl. Frm Table 13.1, we write the standard reductin ptentials as fllws: Cl 2 (1 bar) + 2e 2Cl (a Cl = 1) = V Br 2 (l) + 2e 2Br (a Br = 1) I 2 (s) + 2e 2I (a I = 1) = V = V Applying the diagnal rule, we see that Br2 will xidize I2 but will nt xidize Cl2. Therefre, the nly redx reactin that will ccur appreciably under standard-state cnditins is 5. (10%) Calculate., and ΔG fr the fllwing cell reactins. Assume ideal behavir. (a) Mg(s) + Sn 2+ (aq) Mg 2+ (aq) + Sn(s) [Mg 2+ ] = 0.045M, [Sn 2+ ] = 0.035M (b) 3Zn(s) + 2Cr 3+ (aq) 3Zn 2+ (aq) + 2Cr(s) [Cr 3+ ] = 0.010M, [Zn 2+ ] = M Sl. Strategy: The standard emf ( ) can be calculated using the standard reductin ptentials in Table 13.1 f the text. Because the reactins are nt run under standard-state cnditins (cncentratins are nt 1 M), we need Nernst's equatin [quatin f the text] t calculate the emf ( ) f a hypthetical galvanic cell. Remember that slids d nt appear in the reactin qutient (Q) term in the Nernst equatin. We can calculate G

4 frm using quatin 13.2 f the text: G = nf cell. cell 0.14 V (2.37 V) 2.23 V Frm quatin (13.10) f the text, we write: V lnq n V ln [Mg2 ] n [Sn 2 ] 2.23 V V ln V We can nw find the free energy change at the given cncentratins using quatin 13.2 f the text. Nte that in this reactin, n = 2. G = nf cell G = (2)(96500 J V 1 ml 1 )(2.23 V) = 430 kj ml 1 cell cathde ande Cr 3 /Cr Zn 2 /Zn cell 0.74 V (0.76 V) 0.02 V Frm quatin (13.10) f the text, we write: V lnq n V n ln [Zn2 ] 3 [Cr 3 ] 2

5 0.02 V V ln (0.0085)3 6 (0.010) V We can nw find the free energy change at the given cncentratins using quatin 13.2 f the text. Nte that in this reactin, n = 6. G nf cell G (6)(96500 J V 1 ml 1 )(0.04 V) 23 kj ml 1 6. (10%) Under standard-state cnditins, what spntaneus reactin will ccur in aqueus slutin amng the ins Ce 4+, Ce 3+, Fe 3+, and Fe 2+? Calculate ΔG and Kc fr the reactin. Sl. The half-reactins are: Fe 3 (aq) e Fe 2 (aq) ande 0.77 V Ce 4 (aq) e Ce 3 (aq) cathde 1.61 V Thus, Ce 4 will xidize Fe 2 t Fe 3 ; this makes the Fe 2 /Fe 3 half-reactin the ande. The standard cell emf is fund using quatin (13.1) f the text. cell cathde ande 1.61 V 0.77 V 0.84 V The values f G and Kc are fund using quatins 13.3 and 13.5 f the text. G nf cell (1)(96500 J V 1 ml 1 )(0.84 V) 81 kj ml -1 ln K n cell V n cell (1)(0.84 V) K c = e V e V (10%) A galvanic cell cnsists f a silver electrde in cntact with 346 ml f M AgNO3 slutin and a magnesium electrde in cntact with 288 ml f M Mg(NO3)2 slutin. (a) Calculate fr the cell at 25 C. (b) A current is drawn frm the cell until 1.20 g f silver has been depsited at the silver electrde. Calculate fr the cell at this stage f peratin. (mlar mass f

6 silver = g ml -1 ) Sl. (a) If this were a standard cell, the cncentratins wuld all be 1.00 M, and the vltage wuld just be the standard emf calculated frm Table 13.1 f the text. Since cell emf's depend n the cncentratins f the reactants and prducts, we must use the Nernst equatin [quatin (13.10) f the text] t find the emf f a nnstandard cell V lnq n 3.17 V V ln [Mg2 ] 2 [Ag ] V V ln [0.10] V (b) First we calculate the cncentratin f silver in remaining in slutin after the depsitin f 1.20 g f silver metal ml Ag 1 L 2 Ag riginally in slutin: L ml Ag 1 ml 1.20 g Ag ml Ag g 2 Ag depsited: Ag remaining in slutin: ( ml Ag) ( ml Ag) ml Ag ml 2 [Ag ] M L The verall reactin is: Mg(s) 2Ag (aq) Mg 2 (aq) 2Ag(s) We use the balanced equatin t find the amunt f magnesium metal suffering xidatin and disslving. 2 1 ml Mg 3 ( ml Ag) ml Mg 2 ml Ag The amunt f magnesium riginally in slutin was ml L ml 1 L The new magnesium in cncentratin is: The new cell emf is: 3 2 [( ) ( )]ml L M

7 V lnq n 3.17 V V ln ( ) V 8. (10%) Calculate the pressure f H2 (in bar) required t maintain equilibrium with respect t the fllwing reactin at 25 C: Pb(s) + 2H + (aq) Pb 2+ (aq) + H 2 (g) given that [Pb 2+ ] = M and the slutin is buffered at ph sl. ph 1.60 [H ] M RT 2 [Pb ]PH2 nf ln [H ] V (0.035) P ln H (0.035) P ln H 2 2 P H atm 2 9. (5%) Based n the fllwing standard reductin ptentials: Fe 2+ (aq) + 2e Fe(s) 1 = 0.44V Fe 3+ (aq) + e Fe 2+ (aq) 2 = 0.77V calculate the standard reductin ptential fr the half-reactin Fe 3+ (aq) + 3e Fe(s) 3 =?

8 Sl. It might appear that because the sum f the first tw half-reactins yields the third half-reatin, 3 is given by V. This is nt the case, hwever, because emf is nt an extensive prperty. Because the number f electrns in the first tw half-reactins are nt the same we cannt set On the ther hand, the Gibbs energy is an extensive prperty, s we can add the separate Gibbs energy changes t btain the verall Gibbs energy change. G3 G1 G 2 Substituting the relatinship G = nf, we btain n 3 F 3 n 1 F 1 n 2 F 2 n1 = 2, n2 = 1, and n3 = n 1 1 n2 2 n 3 (2)(0.44 V) (1)(0.77 V) V 10. (10%) Calculate the emf f the Daniell cell at 25 C when the cncentratins f CuSO4 and ZnSO4 are 0.50 M and 0.10 M respectively. What wuld be the emf if activities were used instead f cncentratins? (The γ ± fr CuSO4 and ZnSO4 at their respective cncentratins are and 0.15, respectively.) Sl. A schematic f the Daniell cell is shwn in Figure We need t start by determining the standard ptential fr this cell. cell 0.34 V (0.76 V) 1.10 V Nw we can use the Nernst equatin [see quatin 13.9 in the text] t calculate the ptential at nn-standard cncentratins V lnq n

9 V ln [Zn2 ] n [Cu 2 ] 1.10ÊV V 2 ln 0.10M 0.50M 1.12ÊV If we use activities instead f cncentratins, we get the fllwing result: 1.10ÊV V 2 ln 0.15M 0.068M 1.09ÊV 11. (10%) Predict whether the fllwing reactins wuld ccur spntaneusly in aqueus slutin at 25 C. Assume all species are in their standard states. (a) Ca(s) + Cd 2+ (aq) Ca 2+ (aq) + Cd(s) (b) 2Br(aq) + Sn 2+ (aq) Br 2 (l) + Sn(s) (c) 2Ag(s) + Ni 2+ (aq) 2Ag + (aq) + Ni(s) (d) Cu + (aq) + Fe 3+ (aq) Cu 2+ (aq) + Fe 2+ (aq) Sl. Strategy: cell is psitive fr a spntaneus reactin. In each case, we can calculate the standard cell emf frm the ptentials fr the tw half-reactins. cell = cathde ande Slutin: (a) = 0.40 V (2.87 V) = 2.47 V. The reactin is spntaneus. (b) = 0.14 V 1.07 V = 1.21 V. The reactin is nt spntaneus. (c) = 0.25 V 0.80 V = 1.05 V. The reactin is nt spntaneus. (d) = 0.77 V 0.15 V = 0.62 V. The reactin is spntaneus. 12. (10%) Cnsider a Daniell cell perating under nn standard-state cnditins. Suppse that the cell reactin is multiplied by 2. What effect des this have n each f the fllwing quantities in the Nernst equatin: (a), (b), (c) Q, (d) ln Q, and (e) n? Sl. (a) unchanged (b) unchanged (c) squared (d) dubled (e) dubled

10 13. (5%) Cnsider the galvanic cell as fllwing figure. In a certain experiment, the emf () f the cell is fund t be 0.54 V at 25 C. Suppse that [Zn 2+ ] = M and P H2 = 1.00 bar. Calculate the mlar cncentratin f H +, assuming ideal slutin behavir. Sl. Strategy The equatin that relates standard emf and nnstandard emf is the Nernst equatin. Assuming ideal slutin behavir, the standard states f Zn 2+ and H + are 1 M slutins. The verall cell reactin is Zn(s) + 2H + (? M) Zn 2+ (1. 00M) + H 2 (1. 00bar) Given the emf f the cell (), we apply the Nernst equatin t slve fr [H + ]. Nte that 2 mles f electrns are transferred per mle f reactin, that is, n =2. Slutin As we saw earlier (page 675), the standard emf ( ) fr the cell is 0.76 V. Because the system is at 25 C (298 K) we can use the frm f the Nernst equatin given in quatin 13.5:

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