Chapter 6 6.6 Write the Nernst equation for? - Valdosta State ...

Chapter 66.6 Write the Nernst equation for?(a) The reduction of 0 2 ?In general terms, the Nernst equation is given by the formula:o ⎛ RT ⎞E = E − ⎜ ⎟ln Q⎝ NF ⎠where Q is the reaction quotient. For the reduction of oxygen:O 2 (g) + 4 H + (aq) + 4 e- à 2 H 2 O(1)Q =pOE = Eo21[H+]4⎛ 0.059V− ⎜⎝ 2⎞ ⎛⎟log⎜⎠ ⎝ pO21[H+]4⎞⎟⎠Therefore, the potential for O 2 reduction at pH = 7 and p(O 2 ) = 0.20 bar isE = 1.229 V - 0.42 = 0.81 V(b) The reduction of Fe 2 O 3 (s)?For the reduction of solid iron(III) oxide:Fe 2 O 3 (s) + 6 H + (aq) + 6 e - à 2 Fe(s) + 3 H 2 O(l)1Q =+ 6[ H ]E = Eo⎛ 0.059V⎞ ⎛ 1− ⎜ ⎟log⎜+⎝ 6 ⎠ ⎝[H ]since pH = -log[H + ] = -2.3ln[H + ] and log[H + ] 6 = 6log[H + ].Q1[ H ]=+ 6E = Eo⎛ 0.059V⎞− ⎜ ⎟13.8(pH )⎝ 6 ⎠6⎞⎟⎠

6.8 Using Frost diagrams?(a) What happens when Cl 2 is dissolved in aqueous basic solution?The Frost diagram for chlorine in basic solution is shown in Figure 6.16 and isreproduced below.If the points for Cl - and ClO 4 - are connected by a straight line, Cl 2 lies above it.Therefore, Cl 2 is thermodynamically susceptible to disproportionation to Cl - and C1O 4-when it is dissolved in aqueous base. In practice, the oxidation of ClO - is slow, so asolution of Cl - and ClO - is formed when Cl 2 is dissolved in aqueous base.(b) What happens when Cl 2 is dissolved in aqueous acid solution?The Frost diagram for chlorine in acidic solution is shown in Figure 6.16. If the points forC1 - and any positive oxidation state of chlorine are connected by a straight line, the pointfor Cl 2 lies below it (if only slightly). Therefore, Cl 2 will not disproportionate. However,E° for the C1 2 /C1 - couple, 1.36 V, is more positive than E° for the O 2 /H 2 O couple, 1.23V.Therefore, Cl 2 is thermodynamically capable of oxidizing water as follows, although thereaction is very slow:C1 2 (aq) + H 2 O(l) à 2C1(aq) + 2H + (aq) + 1/2 O 2 (1)E° = 0.13 V(c) Should HClO 3 disproportionate in aqueous acid solution?The point for C1O 3 - in acidic solution on the Frost diagram in Figure 6.16 lies above thesingle straight line connecting the points for Cl 2 and ClO 4 - . Therefore, since C1O 3 - isthermodynamically unstable with respect to disproportionation in acidic solution (i.e. itshould disproportionate), the failure of it to exhibit any observable disproportionationmust be due to a kinetic barrier.

6.9 Write equations for the following reactions:(a) N 2 0 is bubbled into aqueous NaOH solution?Part of the Frost diagram for nitrogen in basic solution is shown below (the entirediagram is shown in Figure 6.9Inspection of it shows that N 2 O lies above the line connecting N 2 and NO 3 - . Therefore,N 2 O is thermodynamically susceptible to disproportionation to N 2 and NO 3 - in basicsolution:5 NO(aq) + 2 OH - (aq) à 2 N0 3 - (aq) + 4 N 2 (g) + H 2 O(l)However, the redox reactions of nitrogen oxides and oxyanions are generally slow. Inpractice, N 2 O has been found to be inert.(b) Zinc metal is added to aqueous acidic sodium triiodide?The overall reaction is:Zn(s) + I 3 - (aq) à Zn 2+ (aq) + 3 I - (aq)Since E° values for the Zn 2+ /Zn and I 3 - /I - couples are -0.76 V and 0.54 V, respectively(these potentials are given in Appendix 2), E° for the net reaction above is:0.54 V + 0.76 V = 1.30 V.Since the net potential is positive, this is a favorable reaction, and it should be kineticallyfacile if the zinc metal is finely divided and thus well exposed to the solution.

(c) I 2 is added to excess aqueous acidic HC1O 3 ?Since E° values for the I 2 and ClO 3 - /C1O 4 - couples are 0.54 V and -1.20 V, respectively(see Appendix 2), the net reaction involving the reduction of I 2 to I - and the oxidation ofC1O 3 - to ClO 4 - will have a net negative potential:E° = 0.54 V + (-1.20 V) = -0.66 V.Therefore, this net reaction will not occur. However, since E° values for the IO - 3 /I 2 andC1O - 3 /Cl - couples are 1.19 V and 1.47 V, respectively, the following net reaction willoccur, with a net E° = 0.28 V:half-reaction: ClO - 3 (aq) + 6 H + (aq) + 6 e - à C1 - (aq) + 3 H 2 O(l)half-reaction: I 2 (s) + 6 H 2 O(l) à 2IO - 3 (aq) + 12H + (aq) + 10e -net rxn: 3I 2 (s) + 5 ClO - 3 (aq) + 3 H 2 0(l) à 6 IO - 3 (aq) + 5 Cl - (aq) + 6 H + (aq)6.10 Will acid or base most favor the following half-reactions?(a) Mn 2+ à Mn0 4 - ?You can answer questions such as this by applying Le Chatelier’s principle to thecomplete, balanced half-reaction, which in this case is:Mn 2+ (aq) + 4 H 2 O(l) à Mn0 4 - (aq) + 8 H + (aq) + 5 e -Since hydrogen ions are produced by this oxidation half-reaction, raising the pH willfavor the reaction. (Alternatively, you could come to the same conclusion by using theNernst equation.) Thus, this reaction is favored in basic solution. At a sufficiently highpH, Mn2 will precipitate as Mn(OH) 2 . The rate of oxidation for this solid will be slowerthan for dissolved species.(b) ClO 4 - à ClO 3 - ?The balanced half-reaction is:ClO 4 - (aq) + 2 H + (aq) + 2 e - à ClO 3 - (aq) + H 2 O(l)Since hydrogen ions are consumed by this reduction half-reaction, lowering the pH willfavor the reaction. Thus, this reaction is favored in acidic solution. This is the reason thatperchioric acid is a dangerous oxidizing agent, even though salts of the perchlorate ionare frequently stable in neutral or basic solution.(c) H 2 O 2 à 0 2 ?The balanced half-reaction is:H 2 O 2 (aq) à O 2 (g) + 2H + (aq) + 2e -Since hydrogen ions are produced by this oxidation half-reaction, raising the pH willfavor the reaction. Thus, the reaction is favored in basic solution. This means thathydrogen peroxide is a better reducing agent in base than in acid.

(d) I 2 à 2I - ?Since protons are neither consumed nor produced in this reduction half-reaction, andsince I is not protonated in aqueous solution (because HI is a very strong acid), thisreaction has the same potential in acidic or basic solution, 0.535 V (this can be confirmedby consulting Appendix 2).6.12 Determine the standard potential for the reduction of ClO 4 - to Cl 2 ?The Latimer diagram for chlorine in acidic solution is given in Appendix 2 and therelevant portion of it is reproduced below:To determine the potential for any couple, you must calculate the weighted average of thepotentials of intervening couples. In general terms it is:and in this specific case it is:0 00( n E + n E + ... + n E )11( n + n + ... + n )1((2)(1.201V) + (2)(1.181V) + (2)(1.674V) + (1)(1.674V))( 2 + 2 + 2 + 1)222nnn== 1.392VThus, the standard potential for the ClO 4 - /Cl 2 couple is 1.392 V. The balanced halfreactionfor this reduction is:2ClO 4 - (aq) + 16H + (aq) + 14e - à Cl 2 (g) + 8H 2 0(l)Note that the point for ClO 4 - at pH 0 in the Frost diagram shown in Figure 6.16 has a yvalue of approximately 9.7, which is NE°/V or 7(1.392).