12-8 Taking the oxidation state of H as + 1 and that of O as - 2, the order of increasing oxidation state of the carbon atoms is

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1 12-1 idation occurs when an atom or molecule loses electrons in the course of a reaction. Reduction occurs when an atom or molecule gains electrons in the course of a reaction (a) The Zn 2+ is reduced and the Cu is oidized. (b) Both the Zn 2+ and Cu 2+ are being reduced and there is no oidization. This cannot occur. (c) The Ag + is reduced and the Zn is oidized. (d) Both the Al 3+ and Ag + are being reduced and there is no oidization. This cannot occur Taking the oidation state of H as + 1 and that of as - 2, the order of increasing oidation state of the carbon atoms is CH 4 (-4) < H 3 CH (-2) < C and H 2 C (0) < C (+2) < C 2 (+4) (a) Na 2 S -2 (e) SF 6 +6 (b) MgS -2 (f) BaS 3 +4 (c) CS 2-2 (g) NaHS 4 +6 (d) H 2 S 4 +6 (h) SCl (a) The C= bond in CH 3 CCH 3 is reduced and the aluminum in LiAlH 4 is oidized. (b) The CH 3 carbon in CH 3 CH 2 H is oidized and the other is reduced. (c) Not a redo reaction. H H -1-1 C C a C a C fumarate H H 0-2 C C C a C H H malate (a) Al(s) + Cr 3+ (aq) Al3+ (aq) + Cr(s) (b) 2 Fe 2+ (aq) + I 2 (aq) 2 Fe3+ (aq) + 2 I - (aq) (c) Cr(s) + 3 Fe 3+ (aq) Cr3+ (aq) + 3 Fe 2+ (aq) (d) Zn(s) + 2 (aq) Zn2+ (aq) + H 2 (g) (a) Al(s) Al3+ (aq) + 3 e 3 e + Cr 3+ (aq) Cr(s) (b) 2 Fe 2+ (aq) 2 Fe3+ (aq) + 2 e 2 e + I 2 (aq) 2 I - (aq) (c) Cr(s) Cr3+ (aq) e 3 e + 3 Fe 3+ (aq) 3 Fe2+ (aq) (d) Zn(s) Zn2+ (aq) + 2 e 2 e + 2 (aq) H 2 (g)

2 12-36 C ( 4) is a reducing agent N (+5) is an oidizing agent S (+2) can act as both Ge (+4) is an oidizing agent As (+3) can act as both Sb (+5) is an oidizing agent N ( 3) is a reducing agent I (+7) is an oidizing agent The better reducing agents are found at the top of Table Manganese, response (b), is the strongest reducing agent of the group (a) The reverse of this reaction should occur. So the reaction will not occur as written. (b) This reaction should occur as written. (c) This reaction should occur as written The given standard-state half-cell reduction potentials indicate that zinc metal is the stronger reducing agent and that copper (+2) is the stronger oidizing agent E represents a potential measured under some general condition. E represents the potential measured for a system under very specific standard conditions Standard conditions for electrochemical measurements have been defined as 1.0 M concentration for solutions and a pressure of 1.0 atm for all gases. Although standard-state measurements can be made at any temperature, they are often taken at 25 C The standard potential for a reaction measures the tendency of the reaction to proceed to the right The magnitude of E for an oidation-reduction reaction indicates the tendency for a given reaction to proceed in the direction described. The potential developed by the cell reflects the relative strengths of a pair of oidizing agents and the relative strengths of a pair of reducing agents. nly the potential difference can be measured The sign of the cell potential tells the direction in which the reaction must shift to reach equilibrium. idation-reduction reactions that have a positive overall cell potential are spontaneous as written. Those that have a negative overall cell potential are not spontaneous as written. An overall cell potential of zero indicates a system at equilibrium The sign of the standard state potential for an oidation-reduction reaction must change when the direction in which the reaction is written is reversed. The magnitude remains the same idation-reduction reactions should occur as written (are spontaneous) when the corresponding cell potential is positive. Reactions (a) and (b) should occur as written, reactions (c) and (d) should occur in the reverse direction E overall = E o + E red (a) E overall = V = V (b) V V = 0.62 V Reaction (b) will occur as written. Reaction (a) will not When a lead acid battery is discharged Pb 2 is reduced to PbS 4. When it is charged up again the reverse reaction is driven through electrolysis.

3 12-78 The dry cell battery, as its name implies, has a lot less water present in the electrolyte solution. The ability to make the cell a lot smaller makes this kind of battery a lot more useful for smaller applications and devices A fuel cell is an electrochemical cell in which a controlled oidation of a fuel produces an electrical voltage in the same way a battery does E cell = E ln Qc n E cell = E o + E red = V + ( V) = V E cell = V = V nfe ln ln at T=298K, K= e RT, where n=3, K= e 196 = This reaction will not proceed. ( Fe ) 3 + ( Al ) = V ne At equilibrium, ln K= = (2)(1.10) = 85.7 K = e 85.7 = The very high value of the equilibrium constant, showing a very large driving force for product formation, justifies the use of the single arrow in the reaction equation An electrolytic cell could not run a motor since electricity must be used to run an electrolytic cell. A voltaic cell can be used to drive a motor, or similarly a voltaic cell can drive an electrolytic cell h 3600 s 1 h = g H amp amp s C 1 mol H 2 2 mol e g H 2 1 mol H h 3600 s 1 h = g 2 mass 2 mass H 2 = g g = amp amp s C 1 mol 2 4 mol e g 2 1 mol 2

4 The number of mols of electrons is 16.5 h 3600 s 1.00 amp = mol e- 1 h amp s C The mol Ce = 21.6 g Ce 1 mol Ce g Ce = mol Ce mol Ce Ratio mol e = = 0.25 mol e mol Ce = = 4 Therefore, for every mol Ce there are 4 mol e - and the cerium ion has an oidation state of 4 +. The formula is CeCl 4, = 1, y = (a)6 I - (aq) + 2 Cr 4 2- (aq) + 16 (aq) 3 I 2 (aq) + 2 Cr 3+ (aq) + 8 H 2 (l) (b)6 Fe 2+ (aq) + 14 (aq) + Cr (aq) 6 Fe 3+ (aq) + 2 Cr 3+ (aq) + 7 H 2 (l) (c)24 H 2 S(g) + 16 Cr 4 2- (aq) + 80 (aq) 3 S 8 (s) + 16 Cr 3+ (aq) + 64 H 2 (l) (a) For the oidation half-reaction: 5(H 2 (l)+ch 3 H (aq) C 2 (g) +6 (aq) +6e - ) For the reduction half-reaction: 6(5e - +8 (aq)+mn - 4 (aq) Mn 2+ (aq)+4h 2 (l)) verall reaction: 5CH 3 H (aq)+18 (aq)+6mn - 4 (aq) 5C 2 (g)+6mn 2+ (aq)+ 19H 2 (l) (b) For the oidation half-reaction: 3(4H - (aq)+ch 3 H (aq) HC 2 H(aq)+3H 2 (l)+4e - ) For the reduction half-reaction: 4(3e - +2H 2 (l) +Mn - 4 (aq) Mn 2 (s) + 4H - (aq) ) verall reaction: 3CH 3 H (aq) + 4Mn - 4 (aq) 3HC 2 H(aq) + H 2 (l) + 4Mn 2 (s) + 4H - (aq) (a) CH 3 CH is the oidized species. (b) CH 3 CCH 3 is the oidized species. (c) Both have the same oidation state. (d) C is the oidized species.

5 e - salt bridge (a) and (b) Ni(s) Pt(s) Ni 2+ 2N 3 - anode Pt 2+ 2N cathode (c) No because the resulting E cell = 0.21V. (d) The nickel will dissolve in the solution and platinum metal will form (a) Ag (b) E cell =1.56 V (c) The Zn cell is the anode and the Ag cell is the cathode. (d) Electrons flow from Zn to Ag. (e) 2 Ag + (aq) + Zn(s) 2 Ag(s) + Zn 2+ (aq) nfe (f) at T=298K, K= e RT, where n=2, K= e = Fe 2+ (aq) + Mn 4 - (aq) + 8 (aq) Mn 2+ (aq) + 4 H 2 (l) + 5Fe 3+ (aq) 3.2 ml Mn mol Mn ml 5 mol Fe mol Mn 4 1 mol FeCl 2 1 mol Fe g FeCl 2 1 mol FeCl 2 = 6.1 g FeCl 2

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