Chem. 6C Midterm 1 Version B October 19, 2007

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1 hem. 6 Miderm Verin Ocber 9, 007 Name Suden Number ll wr mu be hwn n he exam fr parial credi. Pin will be aen ff fr incrrec r n uni. Nn graphing calcular and ne hand wrien 5 ne card are allwed. Prblem (f pible) Prblem (f 9 pible) Prblem (f 8 pible) Prblem 4 (f 0 pible) Prblem 5 (f 0 pible) Prblem 6 (f 5 pible) Prblem 7 (f 6 pible) Prblem 8 (f 5 pible) Prblem 9 (f 6 pible) Miderm Tal (f 50 pible) I wuld lie my grade be ped n line by my uden number

2 ) Mlen magneium chlride i elecrlyzed uing iner elecrde and reacin repreened by he fllwing equain ccur: l - (aq) l (g) +e - Mg + (aq) + e - Mg() ( p) ncerning hi elecrlyi, which f he fllwing aemen i TRUE? a) Oxidain ccur a he cahde. b) Mg + in are reduced a he ande. c) Elecrn pa rugh he meallic par f he circui frm Mg + in he l - in. d) l - in are xidizing agen. e) The cain in he elecrlye underg reducin (nwer) (8 p) Wha curren i required prduce gram f Mg meal in 4 hur? gram Mg ( Mg/ 4. g Mg) 0.08 Mg 0.08 Mg ( e-/ Mg).6 e- 4 h (60 m / h) (60 / m) n I/F rearrange I nf/ ((.6 e-)( )/(50400 )) 0. (0 p) Wha vlume f l (g) i prduced if.00 are paed hrugh he luin fr 4.00 hur a a preure f am a 5º? 4 h (60 m / h) (60 / m) 4400 Q I (.00 )(4400 ) 8800 n Q/F 8800 / f e e- ( l / e-) 0.5 e- am 05 Pa 5º 98.5 K PVnRT rearrange V nrt/p V ((.5 e - )(8.45 J - K - )( 98.5 K))/(05 Pa).70 - m ) The rae f frmain f NO in he reacin N O 5 (g) 4NO (g) + O - i 5.78 ( NO ) - ( p) Wha i he unique average reacin rae? - ¼(5.78).45 - ( p) Wha i he rae f frmain f O? - ¼(5.78).45 ( O ) - ( p) Wha i he rae f N O 5 decmpiin? - ½ (5.78).89 ( N O 5 ) -

3 ) nider he reacin ++ D + E Where he rae law i defined a n experimen i carried u where.00 - and (5 p) fer.00 minue, Wha i he value f in - - ( ) m m d d (8 p) Wha i he half life fr hi experimen? (5 p) Wha i he cncenrain f afer 0.0 minue? The cncenrain f i n effeced by ime herefre he cncenrain f afer 0 m will remain - 4) (0 p) Wrie he cell diagram fr he reacin r + (aq) + H (g) + H + (aq) + r + (aq) P H (g) H + (aq) r + (aq),r + (aq) P()

4 5) The fllwing daa wa bained experimenally a 5º ( - ) ( - ) ( - ) Iniial Rae ( - - ) (5 p)wha i he rae law? Rae x y z x y z x y z 0. (0.) x ln(0.) xln(0.) x y z y z 0.5 (0.50)(.00) y (0.50) z 0.5 (.00) y (0.50) z ln(0.50) yln(.00) + zln(0.50) y (ln(0.50) - zln(0.50))/ln(.00) z y z y z 0.89 (0.5)(0.67) y (.00) z.56 (0.67) y (.00) z ln(.56) yln(0.67) + zln(.00) ( z)ln(0.67) + zln(.00) z z y z y Rae (5 p) Wha i he rae cnan?

5 6) (5 p) alance he fllwing equain: Zn() + NO - (aq) Zn(OH) 4 - (aq) + NH () (baic luin) - Zn Zn(OH) 4 NO - NH - Zn Zn(OH) 4 NO - NH - Zn + H O Zn(OH) 4 NO - NH + H O Zn + 4H O + 4OH - Zn(OH) H O NO - + 9H O NH + H O + 9OH - Zn + 4OH - - Zn(OH) 4 + e - NO - + 6H O + 8e - NH + 9OH - 4(Zn + 4OH - - Zn(OH) 4 + e - ) NO - + 6H O + 8e - NH + 9OH - 4Zn() + 7OH - (aq) + NO - (aq) +6H O(l) 4Zn(OH) - 4 (aq) + NH () 7) ( p) Which f he fllwing i rue fr a galvanic cell?. The elecrn flw i frm negaive elecrde he piive elecrde. he elecrn flw i frm he ande he cahde. The elecrn flw i frm xidizing agen he reduced agen a. nly b. nly c. nly d. and nly (nwer) e.,, and (4 p) Wha are w cmmn difference beween Galvanic cell and elecrlyic cell? Galvanic Elecrlyic cmparmen cmparmen analye analye andard emp & pre nn andard emp & pre N exernal pwer urce Exernal pwer urce 5

6 8) nwer he quein belw uing he fllwing daa Half Reacin Eº (V) g + (aq) + e - g + (aq).99 u + (aq) + e - u().50 Fe + (aq) + e - Fe + (aq) 0.77 H + (aq) + e - H (g) 0.00 Pb + (aq) + e - Pb() -0.6 Ni + (aq) + e - Ni() -0. l + (aq) + e - l() -.66 a + (aq) + e - a() -.76 ( p) Which i he rnge reducing agen? a() ( p) Which i he rnge xidizing agen? g + (aq) ( p) Will l() dilve in - Hl? Ye ( p) an Pb() reduce g + (aq)? Ye ( p) In he elecrchemical cell Ni() Ni + (aq) u + (aq) u(). Wha i he emf when he cell i a equilibrium? equilibrium he emf 0 V 6

7 9) nwer he fllwing quein uing he given galvanic cell MnO - 4 (aq) + 8H + (aq) + 5e - Mn + (aq) +4H O(l) r O - 7 (aq) + 4H + (aq) + 6e - r + (aq) + 7H O(l) Eº.5 V Eº. V ( p) When curren i allwed flw wha pecie i xidized? r + ( p) Wha i he value f Eº cell?.5 V. V 0.8 V ( p) Wha i he xidain ae f r in r O 7 -? 6+ (5 p) Wha i he balanced equain f hi cell? 6(MnO 4 - (aq) + 8H + (aq) + 5e - Mn + (aq) +4H O(l)) 5(r + (aq) + 7H O(l) r O 7 - (aq) + 4H + (aq) + 6e - ) 6MnO 4 - (aq) +0r + (aq) + H O(l) 6Mn + (aq) + 5r O 7 - (aq) + H + (aq) (4 p) Wha i he value f Q, reacin quien, fr hi cell reacin? Q (Mn + 6 r O 7-5 H + )/(MnO 4-6 r + 0 ) ((0.0) 6 (0.0) 5 (0.00) )/((0.0) 6 (0.40) 0 ) (5 p) Wha i he emf a 5º a read n he digial vlmeer? E Eº - (RT/nF)lnQ 0.8V-((8.5 J K - )(98.5 K))/((0)( ))ln( ) 0.6 V (4 p) Wha i he value f he equilibrium cnan a 5º fr he ne pnaneu cell reacin? lnk nfeº/rt K e nfeº/rt e ((0)( )(0.8 V))/((8.45J - K-)(98.5K))

a. (1) Assume T = 20 ºC = 293 K. Apply Equation 2.22 to find the resistivity of the brass in the disk with

a. (1) Assume T = 20 ºC = 293 K. Apply Equation 2.22 to find the resistivity of the brass in the disk with Aignmen #5 EE7 / Fall 0 / Aignmen Sluin.7 hermal cnducin Cnider bra ally wih an X amic fracin f Zn. Since Zn addiin increae he number f cnducin elecrn, we have cale he final ally reiiviy calculaed frm