Balancing Redox Reactions and Identifying Spontaneous Reactions - Prof. Patricia G. Amatei, Exams of Chemistry

Download Balancing Redox Reactions and Identifying Spontaneous Reactions - Prof. Patricia G. Amatei and more Exams Chemistry in PDF only on Docsity! Chemistry 1036 Ayu,A{6'') Test 4 THIS IS FORM A IFORM B #s in ( ) ] April2T,2006 F : 9.65 x lOa C/mol e S pect roc h em ic a I Fsriqs : co >FN.l :;'n*"t') \/ Name RT : 0.0592 V F NOz-> en > NH3 > SCN-> HzO t OH tD Cl- > I- ,4' 1 t't', Choose the single best answer to each question. 1. (10) Balance the following redox reaction in acidic solution. What is the coefficient of H* in the balanced eouation? ,\ t, '' Srjrt- + 2.8 t r-l . ,., MnOq- -+ 3. l0 t-l SOo2- + Mn2* 4. t4 '1 l, I 1.2 Answer: 4. 14 Assign oxidation numbers: +4 -6 +7 -8 \^( t.- +vzv J .| MnOi -8 +6 -2 ---+ SOo2- raTL + Mn2* Write the two half-reactions and balance electrons: +4 +12 TL +6 Oxidation: SrOr' -) 2SO+-2 + 8e- Each of the 2 S atoms loses 4 e- (total = 8; +2+7 Reduction: MnO+- + 5 e- ----' Mn'- Mn gains 5 e- Electrons lost must equal electrons gained; the reduction step must be multiplied by 4 to get a total of 8 e- gained: Oxidation: 5SzO:2- -> 10SO+-2 + 40e- Multiply by 5 for a total of 40 e- Reduction: 8MnO+- + 40 e- ---) 8 Mn'- Multiply by 8 for a total of 40 e- Add the trvo half-reactions: the 40e- sain and the 40e- loss cancel. 5SzO:2- + 8MnO+- ---) l0 SOq2- + 8 Mn2* For acidic solution: l. Add HzO to balance oxygen: There are 47 oxygen atoms on the left side of the equation and 40 oxygen atoms on the right side of the equation. Add 7 H2O molecules to the right side to balance oxygen: 5 SzO:2- + 8 MnO+ ---) 10 SO+2- + 8Mn2* + THzO 2. Add H* to balance hydrogen: there are 14 hydrogen atoms on the right side and 0 hydrogen atoms on the right side. Put 14 H" ions on the left side to balance hydrogen: l4H* + 5 SzO:2- + 8 MnO+- --) l0 So+2- + 8Mn2* + THzO 2.(11) What is the reducing agent in the reaction in the previous question? l. Szo:?- 2. Mno+- 3. So+2- 4. Mn2* Answer: l. SzOr2- The reducing agent is the reactant that undergoes oxidation in the reaction, thus enabling the other reactant to be reduced. In the equation above, SzOr'- is oxidized (reducing agent) while MnO+- is reduced (oxidizing agent). 3.(6) Which of the following statements best describes what will happen when liquid Br2 is poured into a beaker containing aluminum metal? l. Brz will be reduced and Al will be oxidized. 2. Brz will be oxidized and Al will be reduced. 3. Brz will function as the reducing agent and Al will function as the oxidizing agent. 4. A reaction will not occur. Answer: 1. Br' will be reduced and Al will be oxidized. The half reactions involving Br2 and Al are: B1z + 2e' --+ zBf + 1.07 V Al" + 3e- ---+ Al -1.66 V Reverse this since we have Al as a reactant Al --+ Al3" + 3e- +1.66V ox Fe -+ Fez* + 2e- Fe -+ Fe2* + 2e- n:2 e- .-r_ products _ [Fer'l "- r.uctuntr - tAgJT Remember that solid substances are not included in the equilibrium expression The [Ag.] is squared because of the coefficient of 2 inthe balanced equation. E = t.24 v - 0'0592 v lon [l'50] - 2 e- - [0.0s0]' E= 1.16V 8.(12) A current of 2.50 A was passed through an electrolysis cell containing molten CaClz for 4.50 hours. How many grams of calcium metal are deposited? 1. 16.89 2. 2.339 3. 8.429 4. ll,2g Answer: 3. 8.42 g Current x time -> coulombs -> moles electrons -> mole metal -+ mass metal Caz* + 2 e- ---) Ca(s) 2 moles of e- needed for every I mole of Ca produced ^.(4.s0h 3600s)C:Axs C=2.50A1 | C:40500C\ th / 40500 C I mol e- I mol Ca 40.1 s, X ---------------- X --------: X ---'-i--e-:8,42g 9.65 x l0* C 2 mol e- I mol Ca 9.(3) A voltaic cell is composed of the following two half-cells: Co/Co2* and Fe/Fe"-. Which of the following statements is true? l. Co2* is being produced at the cathode. 2. Fe2* is being produced at the cathode. 3. The anode is made of Fe. 4. The reaction is not spontaneous. Answer: 3. The anode is made of Fe. Since this is a voltaic cell, we have to write a spontaneous reaction: ^ )+.Co^" (aq) + 2e- ----+ Co(s) -0.28 V Fe'-(aq) + 2e- --+ Fe(s) -0.44V Since Fe2+ is below Co2+ in the half-cell potential table, the Fe2+ reaction will spontaneously pe the oxidation: Reduction: Co'-(aq) + 2e' ---) Co(s) -0.28 V Reduction occurs at the cathode; Co(s) is being produced at the cathode, not Co2*(aq;. Oxidation occurs at the anode. The anode will be made of Fe(s) since Fe is undergoing oxidation. 10.(7) Consider the following voltaic cell. Calculate AGo, the standard free energy at 25"C. 2 Cr(s) + 3 Sn2*(aq) -+ 2 Cr3*(aq) + 3 Sn(s) 1. 'r74kl 2. -509kJ Answer: 4. -341kJ AGo: -nF Eo 3. -10r2 kJ 4. -347 kI Ox: Cr -+ Cr'- + 3e- 0.74 V (sign has been reversed for the oxidation) Red: Sn2* + 2e- -+ Sn -0.14V Ox: 2Cr --> 2Cr3n +6e- 0.74V Red: 3Sn2* + 6e- -+ 3Sn -0.14 V n :6e- Eo: 0.74V + (-0.14 V) = 0.60 V Oxidatiqn: Fe(s) -+ Fe'-(aq) + 2e- . +0.44 V (note sign change) Overall: Co'-(aq) + Fe(s) ---+ Co(s) + Fe'-(aq) E: -0.28 + 0.44 V:0.16 V AG o: -6mole- (9.65 x 104 C/mole-X0.60 J/C) AGo : - 347400 J: -347 kI I l.(4) Which of the following can be reduced by Ni? l. Mn2n 4. Mn2* andZnz* Answer: 5. Ag* and Pb2* l V: J/C 2. Ag* 5. Ag* and Pb2* 3. Zn2* A substance can reduce any substance above it in the cell potential table and cannot reduce a substance below it in the table. So Ni can reduce the ions above it in the table. Here is the list of the substances in the problem above in the order in which they appear in the Table: Ag Pb2* Ni Znz* Mn2* Since Zn2*and Mn2*occur below Ni in the table, those ions cannot be reduced by Ni.. Since Agn and Pb2" occur above Ni in the table, those ions can be reduced by Ni. 12.(8)What is the value of the equilibrium constant, K. for the reaction below? Sn2*(aq) + Fe(s) -'--' Sn(s) + Fe2*(aq) Eo : +0.30 V l. 1.4 x l0r0 z. 2.4 x 1016 3. 1.2 x 105 4. 4.1x 1020 Answer: l. 1.4 x 10ro Ox: Fe -+ Fe2* + 2e- Red: Sn2*_ + 2e- -+ Sn n :2e- Eo: 0.30 V no 0.0592 V .E": -'--'- losK n 0.0592 V, 0.30 V : "'- los K2- 10.135 : loe K l0ro r35 : K-: 1.4 x lOto Ij -r-'I t-L 13.(16)What is the oxidation number of Fe in K2[Fe(OH)z(HzOXCl)z]? l.+2 2. +3 3.+4 4.0 Answer: 1. +2 First, find the charge on Mn and determine the number of d electrons: +20 [Mn(HzO)o]2*^ Since the water ligand! are neutral, the +2 charge on the complex is due to the Mn2* ion. Mn: [Ar]4s2 3ds Mn2*: [Ar]3ds Put the 4 electrons in the orbitals: _t_ Since there are 5 unpaired electrons, the complex is paramagnetic (having unpaired electrons). Two electrons have crossed the crystal field energy gap to remain unpaired. This will only happen when the crystal field f tf splitting energy is small. Since the number of unpaired electrons is maximized. the complex is said to be high spin. 19.(25)Consider the complex ion [M(NH3)6]* where M is a d6 metal ion. What change is expected when all the NH3 ligands are replaced by F- ligands? 1. The complex changes from having 4 unpaired electrons to 0 unpaired electrons. 2. The complex becomes a different color. 3. The complex becomes low spin. 4. The complex becomes tetrahedral. 5. No change occurs. Answer: 2. The complex becomes a different color. Different ligands cause different amounts of crystalfield splitting energy. F- causes a smaller splitting than NHr. Therefore, a complex with F- ligands will absorb light of lower energy than complexes with NH3 ligands and the two complexes will be different colors. Since NH: causes more splitting than F- ligands, the NH: complex will be low spin and have 0 unpaired electrons and the F- complex will be high spin and have 4 unpaired electrons. Both complexes, regardless of ligand, have a coordination number of 6 and will be octahedral. 20.(24)Consider the following complex ions. Which one is green in color? (M = metal) l. [M(H2O)6]3* 3. [M(en)3]3* Answer: 4. [MI6]3- t_ 2. 1M(Clr)013- 4. [MIo]'- .N A complex that is green in color absorbs the complement color which is red (from the color wheel). In the visible spectrum, red light is lowest in energy. A complex that absorbs low energy light must have a relatively small crystal field splitting energy. The ligand above that causes the smallest crystal field splitting energy is the ligand that lies furthest to the risht in the spectrochemical series. That would be l-. 21.(19)Whi"h o-n; of the following is diamagnetic? +9 -(J'{S(€Ndir-*- 3. [Cr(NH3)6]3* +t o 2. 4. .6tl -l [Co(CN)6]' [Ni(HzO)o]'- ,. J Tl/ rlr r.'lv \ \ 1t,l \y rY Answer: 2. [Co(CN)6]3- A diamagnetic complex has no unpaired electrons. Determine the charge on the metal ion and the number of d electrons: -6 +3 -l 1. 3- lFe(CN)el CN- ligands form low spin complexes; there is one unpaired electron - paramagnetic. (The complex is also paramagnetic if you draw it as high spin). AI A I t.t l.t I -6 +3 -l 3-2. [Co(CN)e] co: [Ar]4s2 3d7 Co3*: 1Ar13d6 CN- ligands always form low spin complexes; allthe electrons are paired: diamagnetic f t f I f It\y I \y t\y +30 3. [CTOIH:)o]3* Cr: [Ar]4sr 3d5 Cr3*: [Ar]3d3 Fe: [Ar] 4s2 3d6 Fe*3: [Ar]3d5 t tt 4. [Ni(H2O)6]2* _1_ r There are three unpaired electrons Ni: [Ar]4s2 3d8 Ni2*: [Ar]3d8 There are two unpaired electrons ^t f | ,ttt\r. | \l/ t\y 22.(13)Which one of the following is the correct electron configuration for the Fe3* ion? 2. lArl4s2 3d3r-----*----- _-T i*L5rlj:j**, Answer: 5. [Ar] 3ds Fe: [Ar] 4s2 3d6 Fe3*: [Ar]3d5 remove the 2 4s electrons and I of the 3d electrons. +3 +3 -b') \ _\ 23.(2})"Cobalt Yellow" is a pigment used in oilpaints, and contains the coordination compound K:[Co(NOz)o]. How many unpaired electrons are there on the cobalt atom in this compound? Note that NOz- is a strong-field ligand. 3. two 4. three 5. four -lr. 4:'tl \1 l 7 Find the oxidation nurnber of Co and the number of d electrons in that ion: +3 +3 -6 +l +3 -l K:[Co(NOz)e] *g'S="i{: 4. [Ar] 3d" 7. zero 2. one Answer: l. zero Co: [Ar] 4s2 3d7 Co3*: [Ar]3d6