UNIT 7 Assignment Electrochemistry (Chap 5-pg & Chap 19-pg ) ANSWERS. 2 MnO 4 2 NO 3

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1 UNIT 7 Assignment Electrochemistry (Chap 5pg & Chap 19pg ) ANSWERS 7.1. Assign Oxidation numbers to the elements in bold print. 1) Mg 0 2) P 4 0 3) Cl I 4) HCl I 5) CaO +II 6) Na 2 O +I 7) Fe 2 O 3 +III 8) AlF 3 +III 9) P 2 O 3 +III 10) PF 3 +III 11) SO 2 +IV 12) PCl 5 +V 13) ThO 2 +IV 14) CO 2 +IV 15) CO+II 2 16) CO 3 +IV 17) SnS 2 +IV 2 18) SO 4 +VI 19) PO 4 3+V 20) AlNIII 21) Fe 3 O 4 8/3 22) H 2 O 2 I 23) H 2 SII 24) ClO +I 25) NaClO 2 +III 26) KClO 3 +V 27) Mg(ClO 4 ) 2 +VII 28)LiMnO 4 +VII 29) K 2 CrO 4 +VI 30) Na 2 Cr 2 O 7 +VI 31) SO 3 2+IV 32) NO+II 33) NO 2 +IV 34) N 2 O 3 +III 35 ) N 2 O 5 +V 36) N 2 H 4 II 37) N ) NH 2 OHI 39) NH 3 III 40) N 2 O+I 41) Ca(VO 3 ) 2 +V 42) MnO 2 +IV 43) CHCl 3 +II 44) C 2 H 5 OHII 45) C 12 H 22 O ) CaCO 3 +IV 47) NaHCO 3 +IV 48) BrCl+I 49) SCl 2 +II 7.2. Balance the following ionic eq'ns taking place in acid solution using oxidation #'s 1) MnO 4 2) 2 MnO 4 3) 2 Cr 2 O 7 4) 2 NO 3 5) 2 SO Fe H + Mn Fe H 2 O + 5 Sn H + 2 Mn Sn H 2 O + 6 Fe H + 2 Cr Fe H 2 O + 3 Cu + 8 H + 3 Cu NO + 4 H 2 O + Cu + 4 H + Cu 2+ + SO H 2 O 6) NO Zn + 10 H + 4 Zn 2+ + NH H 2 O 7) H 2 SO Fe 3+ + H 2 O 2 Fe 2+ + SO H + 8) 2 MnO H 2 C 2 O H + 2 Mn CO H 2 O 9) 4 Fe 3 O H 2 S 12 FeS + SO H H 2 O 10) 2 CeO Cl + 8 H + 2 Ce 3+ + Cl H 2 O 11) 2 NO 3 12) NO 3 13) 2 NO 3 + Cu + 4 H + Cu NO H 2 O + 3 Fe H + 3 Fe 3+ + NO + 2 H 2 O + 5 Zn + 12 H + 5 Zn 2+ + N H 2 O

2 14) 2 NO 3 + H 2 S + 2 H + S + 2 NO H 2 O 15) 4 Fe NH 3 OH + 4 Fe 2+ + N 2 O + 6 H + + H 2 O 16) 2 HNO 2 + C 2 O H + 2 CO NO + 2 H 2 O 17) 2 MnO Br + 8 H + 3 Br MnO H 2 O 18) 3 C 2 H 3 OCl + 4 Cr 2 O H + 8 Cr CO Cl + 19 H 2 O 19) 2 CHCl MnO H + 3 Cl CO Mn H 2 O 20) SO ) MnO I + 10 H + 4 I 2 + H 2 S + 4 H 2 O + 5 Fe 3 O H + 15 Fe 3+ + Mn H 2 O 22) 2 CrO H 2 O OH 2 CrO H 2 O 23) 2 CrO HSnO 2 + H 2 O 2 CrO HSnO OH 24) 2 CrO S 2 O OH 2 CrO SO H 2 O 7.3: Balance the following stoichiometric equations using oxidation numbers 1) 3 Cu + 8 HNO 3 3 Cu(NO 3 ) NO + 4 H 2 O 2) Cu + 4 HNO 3 Cu(NO 3 ) NO H 2 O 3) MnO HCl MnCl 2 + Cl H 2 O 4) Pb 3 O HCl 3 PbCl 2 + Cl H 2 O 5) 2 KMnO HCl 2 KCl + 2 MnCl Cl H 2 O 6) 2 K 2 CrO HBr 4 KBr + 2 CrBr Br H 2 O 7) H 2 S + 2 HNO 3 2 NO 2 + S + 2 H 2 O 8) 2 HNO HCl 2 NO + 3 Cl H 2 O 9) 5 Sb 2 (SO 4 ) KMnO H 2 O 10 H 3 SbO K 2 SO MnSO H 2 SO 4

3 10) 5 CaC 2 O KMnO H 2 SO 4 5 CaSO 4 + K 2 SO MnSO H 2 O + 10 CO 2 11) 10 FeSO KMnO H 2 SO 4 5 Fe 2 (SO 4 ) 3 + K 2 SO MnSO H 2 O 12) KIO KI + 6 HCl 6 KCl + 3 I H 2 O 13) 10 KI + 2 KMnO H 2 SO 4 5 I K 2 SO MnSO H 2 O 14) I HNO 3 2 HIO NO H 2 O 15) K 2 Cr 2 O HCl 2 KCl + 2 CrCl Cl H 2 O 7.4. Drawing electrochemical cells. (19.5 pg 866) On each diagram a) label the anode & cathode b) show the electron flow c) show the moement of all ions in the reactions & the salt bridge. Ag / Ag + //sb// Zn 2+ / Zn Ag / Ag + //sb// Au 3+ / Au A B A B 1/2 rx(a) Zn Zn e E = /2 rx(a) Ag Ag e E = /2 rx(b) Ag e Ag E = /2 rx(b) Au e Au E = 1.50 cell rx Zn + 2Ag + Zn Ag E = 1.56 cell rx 3Ag + Au 3+ 3 Ag + + Au E = 0.70 Al / Al 3+ //sb// Mg 2+ / Mg Cd / Cd 2+ //sb// Ni 2+ / Ni A B A B 1/2 rx(a) Mg Mg e E = /2 rx(a) Cd Cd e E = /2 rx(b) Al e Al E = /2 rx(b) Ni e Ni E = 0.25 cell rx 3Mg + 2Al 3+ 3Mg Al E = 0.71 cell rx Cd + Ni 2+ Cd 2+ + Ni E = 0.15

4 Ag / Ag + //sb// Cl / Cl 2 Cr / Cr 3+ //sb// I / I 2 A B A B 1/2 rx(a) Ag Ag e E = /2 rx(a) Cr Cr e E = /2 rx(b) Cl e 2Cl E = /2 rx(b) I e 2I E = 0.53 cell rx 2Ag + Cl 2 2Ag + + 2Cl E = 0.56 cell rx 2Cr + 3I 2 2Cr I E = 1.27 I 2 / I //sb// Cl / Cl 2 F 2 / F //sb// Li + / Li A B A B 1/2 rx(a) 2I I e E = /2 rx(a) Li Li e E = /2 rx(b) Cl e 2Cl E = /2 rx(b) F e 2F E = 2.87 cell rx 2 I + Cl 2 I 2 + 2Cl E = 0.83 cell rx 2Li + F 2 2Li F E = 5.92 *Practical applications of EC Cells Batteries (19.9 pg 885) 7.5. Attempt questions a) b) & c) on pg a) Pb + SO 4 PbSO 4 + 2e anode PbO H + + SO e PbSO H 2 O cathode Connecting 6 2olt cells in series produces 12 olts b) Zn + 2 OH ZnO + H 2 O + 2e anode 2 MnO 2 + H 2 O + 2e Mn 2 O OH cathode c) Fuel cells are more efficient thermodynamically and more of the energy of the reaction can be made aailable for useful work. As well the energy is now in the form of electricity and can be transported easier, stored more efficiently etc ( read pg )

5 * Calculating Cell Potentials / E alues 7.6. Attempt questions a) b) on pg 893 a) The relatie difference between copper & hydrogen is 0.34 and a constant regardless of the reference point. So if the standard reduction electrode was copper E (Cu2+) =0.00 then hydrogen would hae a alue of 0.34 or E (H+) =0.34 b) The negatie terminal of the oltmeter must be connected to the anode to register any oltage if it is connected backwardsthe oltmeter will continue to read 0.00 or a negatie alue as the current is running in the wrong direction for the meter 7.7. Which of the following substances will oxidize Br to Br 2? a) Cl 2 b) H + c) Ni 2+ d) MnO 4 To oxidize Br to Br 2 the substances must be below it on the EC series Cl 2 & acidified MnO 4 ( a & d ) are below Br 7.8. Which of the following substances will reduce Br 2 to Br? a) Cl b) H 2 c) Ni d) Mn 2+ To reduce Br 2 to Br the substances must be aboe it on the EC series H 2 & Ni ( b & c ) are aboe Br Attempt questions a) b) on pg 894 a) The better reducing agent will be oxidized in any rx so will be higher on the Oxid Pot list or will be lower on the Red Pot list i) Sn ii) Br iii) Zn i) I b) The better oxidizing agent will be reduced in any rx so will be lower on the Oxid Pot list or will be higher on the Red Pot list i) MnO 4 ii) Au 3+ iii) PbO 2 i) HOCl 7.10 Attempt questions a) b) c) d) e) f) g) h) i) j) k) on pg a) Cd 2+ + Fe Cd + Fe 2+ E = 0.40(0.44) = 0.04 Br 2 + 2Cl Cl Br E = 1.07(1.36) = 0.29 Au Ag Au + 3 Ag + E = 1.42(0.80) = 0.62 b) NO H Fe 2+ 3 Fe 3+ + NO + 2 H 2 O E = 0.96(0.77) = 0.19 NiO H + + 2Ag Ni H 2 O + 2 Ag + E = 0.49(0.80) = 0.31 Mg + Cd 2+ Mg 2+ + Cd E = 0.40(2.37) = 1.97 c) 2 Au I 3 I Au E = 1.42(0.54) = (spont) 3 Fe NO + 4 H 2 O 3 Fe + 2NO H + E = 0.44(0.96) = 1.40 (not spont) 3 Ca + 2 Cr 3+ 2 Cr + 3 Ca 2+ E = 0.74(2.76) = (spont)

6 d) MnO H Fe 2+ 5 Fe 3+ + Mn H 2 O E = 1.51(0.77) = 0.74 e) 2 Ag + + Fe 2 Ag + Fe 2+ E = 0.80(0.44) = 1.24 f) net rx = BrO H I 3 I 2 + Br + 3 H 2 O E = 1.44(0.54) = 0.90 (spont) g) net rx = MnO H + + Pb + 2Cl Mn H 2 O + PbCl 2 E = 1.23(0.27) = 1.50 (sp) h) net rx = 4 HOCl + 2 H S 2 O 3 2 Cl 2 + H 2 O + 2 H 2 SO 3 E = 1.63(0.40) = 1.23 (sp) i) net rx = Br I I Br E = 1.07(0.54) = 0.53 (spont) j) net rx = 2 SO H I H 2 SO 3 + I 2 + H 2 O E = 0.17(0.54) = 0.37 (not spont) k) net rx = S 2 O Ni(OH) OH 2 SO NiO H 2 O E = 2.01(0.49) = 1.52 (spont) If a piece of copper metal is dipped into a 1 M Cr 3+ sol'n, what will happen? possible rx = Cu + 2 Cr 3+ Cu Cr 2+ E = 0.75 (not spont) so nothing happens when Cu is dipped into 1 M Cr 3+ sol'n What will happen if an aluminum spoon is used to stir a sol'n of Fe(NO 3 ) 2? What kind of spoon would be better? possible rx = Ni + 2 Fe 3+ Ni Fe 2+ E = 1.02 (spont) so nickel is not a suitable container Can a 1 M Fe 2 (SO 4 ) 3 sol'n be stored in a container made of nickel? possible rx = 2 Al + 3 Fe 2+ 2 Al Fe E = 1.22 (spont) so the ferrous sol'n will oxidize the aluminum spoon For each of the following write the equation for the reaction that is most likely to occur and include E ( all sol'n are 1 M ) a) A mixture of powdered I 2 & liquid Br 2 are placed in a sol'n containing both I & Br. what could happen? I e 2 I E = 0.53 Br e 2 Br E = I I e E = Br Br e E = 1.06 NET RX = Br I 2 Br + I 2 E = 0.53

7 b) A tin strip is immersed in HNO 3 (aq) what could happen? Sn Sn e E = H e H 2 E = NO H e 2 NO + 4 H 2 O E = 0.96 NET RX = 3 Sn + 2 NO H + 2 NO + 4 H 2 O + 3 Sn 2+ E = 1.10 also Sn + 2 H + H 2 + Sn 2+ E = 0.14 c) A mixture of powdered Al and Fe is added to a sol'n of Cr(NO 3 ) 3 what could happen? Cr e Cr 2+ E = 0.41 Cr e Cr E = 0.74 Al Al e E = 1.66 Fe Fe e E = 0.44 NET RX s = Al + 3 Cr 3+ Al Cr 2+ E = 1.25 Al + Cr 3+ Al 3+ + Cr E = 0.92 Fe + 2 Cr 3+ Fe Cr 2+ E = 0.03 d) A copper rod is immersed in HCl (aq) through which is bubbled O 2 (g) What could happen Cu Cu e E = H e H 2 E = /2 O H + H 2 O E = 1.23 NET RX = Cu + 1/2 O H + Cu 2+ + H 2 O E = 0.89 *The Effect of Concentration on Cell Potentials Attempt questions a) b) c) & d) on page a) NiO H Ag Ni H 2 O + 2 Ag + E = 2.48 E = E ( 0.06 / n log 10 ([Ag + ] 2 [ Ni 2+ ] / [H + ] 4 )) nearnst eq n E = / 2 log 10 ([0.01] 2 [ 0.01] / [1 x 10 5 ] 4 )) E = log 10 (( 1 x 10 6 ) / (1 x )) = = 2.07

8 b) 3 I Cr 2 O H + 6 IO Cr H 2 O E = E = E ( 0.06 / n log 10 ([Cr 3+ ] 10 [IO 3 ] 6 / [H + ] 34 [Cr 2 O 7 2 ] 5 )) nearnst eq n E = / 30 log 10 ([1 x 10 3 ] 10 [1 x 10 4 ] 6 / [1 x 10 1 ] 34 [1 x 10 2 ] 5 )) E = log 10 (( 1 x ) / (1 x )) = = c) Mg + Cd 2+ Mg 2+ + Cd E = 1.97 E = E ( 0.06 / n log 10 ([Mg 2+ ] / [Cd 2+ ] ) 1.54 = / 2 log 10 (1 / [Cd 2+ ]) 0.43 = 0.03 log 10 (1 / [Cd 2+ ]) 1 / [Cd 2+ ] = = 2.14 x [Cd 2+ ] = 4.7 x M nearnst eq n d) 2 Ag + 2 Cl 1 + Cu 2+ Cu + 2 AgCl E = E = E ( 0.06 / n log 10 ( 1 / [Cu 2+ ][ Cl 1 ] 2 ) nearnst eq n = / 2 log 10 (1 / [1.0][ Cl 1 ] 2 ) = 0.03 log 10 (1 / [Cl 1 ] 2 ) 1 / [Cl 1 ] 2 = = 8.0 [Cl 1 ] 2 = [Cl 1 ] = M *Cell Potentials and Thermodynamics Attempt questions a) b) & c) d) e) f) on page 895 a) 2 Br + I 2 Br I E = 0.53 G = n F E = 2 * * 0.53 = kj b) 2 MnO H HCHO 2 2 Mn H 2 O + 5 CO 2 E = = 1.69 G = n F E = 10 * * 1.69 = kj

9 c) 5 S 2 O Cl H 2 O 10 SO ClO H + E = 2.01(1.47) = 0.54 G = n F E Ke = 10 ne/0.06 = 10 * * 0.54 = 10 = 521 kj = 1 x *0.54 / 0.06 d) Co + Ni 2+ Co 2+ + Ni E = 0.25(0.28) = 0.03 Ke = 10 ne/0.06 = 10 = 1 x *0.03 / 0.06 e) 2 AgI + Sn Sn Ag + 2 I E = Ke = 10 ne/0.06 2*0.015 / 0.06 = 10 = 1 x = f) 2 H 2 O + 2 Cl 2 4 H Cl + O 2 E 1.36(1.23) = 0.13 Ke = 10 ne/0.06 4*0.13 / 0.06 = 10 = 1 x = 4.7 x 10 8 * Electrolysis Attempt questions a) b) c) d) e) f) g) h) i) j) on page a) Cu 2+ is more readily reduced than K + and water so expect Cu at the cathode Br is more readily oxidized than NO 3 and water so expect Br 2 b& c) ii) Q Q = I * t = 4.00 * 600 s = 2400 coul = Faradays(moles of electrons) = I * t = 10.0 * 20 * 60 s = coul = Faradays iii) Q = I * t = 1.50 * 6 * 60 * 60 s = coul = faradays d) i) 0.2 mol Fe 2+ Fe requires 0.2 * 2 = 0.4 Faradays ii) 0.7 mol Cl Cl 2 requires 0.7 * 1 = 0.7 Faradays iii) 1.5 mol Cr 3+ Cr requires 1.5 * 3 = 4.5 Faradays i) 0.01 mol Mn 2+ MnO 4 requires 0.01 * 5 = 0.05 Faradays

10 e) i) 1 mol Mg 2+ Mg requires 2 Faradays ii) 1 mol Cu 2+ Cu requires 2 F so 24 g Mg 2+ " 2 " so 63.5 g Cu 2+ " 2 " 5 g " " F 41 g " " 1.3 F f) # moles Ag reduced = 12 g / g = mol **since Cr requires 3 F to be reduced and Ag only 1 F # moles of Cr reduced = / 3 = mol g) time = q/i = F * coul/f / 4 amp = x10 3 s = 44.6 min h) time = q/i = 75g/52 *3F * coul/f / 2.25 amp = x10 5 s = 51.5 h i) ) # mol of Al reduced = / 27 = x 10 4 moles # Far req'd = 1.52 x 10 4 * 3 = 4.54 x 10 4 F # coul = 4.54 x 10 4 F * = 4.4 x 10 9 coul I = Q /t = 4.4 x 10 9 coul / 8.64 x 10 4 s = 5.1 x 10 4 amps j) # coul = 1.5 * 30 * 60 s = 2.7 x 10 3 coul # far = 2.7 x 10 3 / = 2.8 x 10 2 F # mol Van reduced = g / g = 9.3 x 10 3 mol so oxid state of Van = 2.8 x 10 2 F / 9.3 x 10 3 mol = 3 or V 3+ V 7.18 ** Summary question on pg MnO H Cl 5 ClO Mn H 2 O E = 1.51(1.45) = G = n F E Ke = 10 ne/0.06 = 30 * * = 10 30*0.056 / 0.06 = 162 kj = 1 x = 2.4 x E = E ( 0.06 / n log 10 ([Mn 2+ ] 6 [ClO 3 ] 5 )/([MnO 4 ] 6 [H + ] 18 [Cl ] 5 ) nearnst eq n E = / 30 log 10 ([0.05] 6 [0.110] 5 ) /([0.20] 6 [5.62x10 5 ] 18 [0.003] 5 ) E = log 10 (( ) / (4.87 x )) = = 0.105

Unit #8, Chapter 10 Outline Electrochemistry and Redox Reactions

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