Skoog Chapter 22 Jawaban [PDF]

Zitiervorschau

Nama : Siti Rahayu NIM

: 1806021

Kelas : A Skoog Chapter 22 Jawaban : 22.1

Calculate the electrode potentials of the following half-cells a. Ag+(0.0261 M)|Ag b. Fe3+(6.72 x 10-4 M), Fe2+ (0.l00 M)|Pt c. AgBr(sat'd), Br' (0.050 M)|Ag Jawab : a)

E = 0.799 =

0.0592 1 log 1 ¿¿

E = 0.799 – 0.0592 log

1 0.0261

E = 0.799 – 0.094 E = 0.705 V b)

E = 0.771 – 0.0592 log

0.100 6.72 x 10⁻ ⁴

E = 0.711 – 0.129 E = 0.642 V c)

E = 0.073 – 0.0592 log 0.05 E = 0.141 V

22.2

Calculate the electrode potentials of the following half-cells. a. HCl(1.76 M) | H2(0.987 atm), Pt b. IO3-(0.194 M), I2(2.00 x 10-4 M), H+ (3.50 X 10-3 M)| Pt c. Ag,CrO.(sat'd),CrO,'-(0.0520 M)IAg Jawab :

a)

E = 0.000 E=-

0.0592 PH ₂ log 2 ¿¿

0.987 0.0592 log 2 (1.76) ²

E = 0.015 V b)

0.0592 (2 x 10−4 )¹ /² E = 1.178 log 5 (0.194) ¿ ¿ E = 0.826 V

c)

E = 0.446 -

0.0592 log 0.0520 2

E = 0.446 + 0.038 E = 0.484 V 22.3

For each the following half-cells, compare the electrode potentials calculated from (1)

concentration and (2) activity data. a. HCl(0.0200 M),NaCl(0.0300M)|H2(1.00 atm)|Pt b. Fe(ClO4)2(0.0111 M),Fe(ClO4)3(0.0111M)|Pt Jawab : a)

2H+ + 2e- ↔ H2 (1) E = 0.000 –

E0 = 0.000 V 1 0.0592 log [ ] = -0.101 V 2 (0.020)²

(2) Kekuatan ion : μ = ⅟₂ [0.02 x 1² + 0.03 x 1² + 0.05 x 1²] = 0.050 Dari tabel a2-1, ϒH+ = 0.86 dan αH+ = 0.86 x 0.02 = 0.0172 E = 0.000 – b)

0.0592 log ¿] = 0.104 V 2

Fe3+ + e- ↔ Fe2+

E0 = 0.771 V

(1) E = 0.771 – 0.0592 log

0.0111 = 0.771 V 0.0111

(2) μ = ⅟₂[0.0111 x 2² + 0.0111 x 3² + 2 x 0.0111 x 1² + 3 x 0.0111 x 1²] μ = 0.100 dari tabel a2-1 ϒFe2+ = 0.40

dan

αFe2+ = 0.04 x 0.0111 = 0.00444

ϒFe3+ = 0.18

dan

αFe3+ = 0.18 x 0.0111 = 0.00200

E = 0.771 – 0.0592 log

22.4

0.00444 = 0.750 V 0.00200

For each the following half-cells, compare the electrode potentials calculated from (1)

concentration and (2) activity data. a. Sn(ClO4)2(3.00 x 10-5 M), Sn(ClO4)4 (6.00 x 10-5 M)|Pt b. SN(ClO4)2(3.00 x 10-5 M), Sn(ClO4)4 (6.00 x 10-5 M), NaClO4 (0.0800 M)|Pt Jawab : a)

Sn2+ + 2e- ↔ Sn2+ (1) E = 0.154 –

E0 = 0.154 V

¿ 0.0592 log [3.00 x 10 ⁻⁵ ¿ −5 ]= 0.163 V (6.00 x 10 ) 2

(2) μ = ⅟₂ [3.00 x 10⁻⁵ x 2² + 6.00 x 10⁻⁵ x 4² + 6.00 x 10⁻⁵ x 1² + 24 x 10⁻⁵ x 1²] = 6.9 x 10⁻⁴ Dari persamaan a2-3 -log ϒSn2+ =

0.509(2)² √ 6.9 x 10 ⁻ ⁴ = 0.05085 1+ 3.28 x 0.6 √ 6.9 x 10⁻ ⁴

ϒSn2+ = 0.890 -log ϒSn4+ =

b)

αSn2+ = 0.890 x 3.00 x 10⁻⁵ = 2.67 x 10⁻⁵

0.509(4 )² √6.9 x 10 ⁻ ⁴ = 0.195 1+ 3.28 x 1.1 √ 6.9 x 10 ⁻ ⁴

ϒSn4+ = 0.638 E = 0.154 -

dan

dan

αSn4+ = 0.638 x 6.00 x 10⁻⁵ = 3.83 x 10⁻⁵

0.0592 (2.67 x 10−5) log [ ]= 0.159 V 2 (3.83 x 10 ⁻⁵)

(1) dari jawaban no 4a, E = 0.163 V (2) disini kontribusi spesi timah untuk μ diabaikan dan μ = 0.0800 -log ϒSn2+ =

0.509 (2)² √ 0.080 = 0.3699 1+ 3.28 x 0.6 √ 0.080

ϒSn2+ = 0.427 -log ϒSn4+ =

dan

αSn2+ = 0.427x 3.00 x 10⁻⁵ = 1.28 x 10⁻⁵

0.509(4 )² √ 0.080 = 1.140 1+ 3.28 x 1.1 √ 0.080

ϒSn4+ = 0.072

dan

αSn4+ = 0.072 x 6.00 x 10⁻⁵ = 4.35 x 10⁻⁶

0.0592 (1.28 x 10−5 ) E = 0.154 log [ ] = 0.140 V 2 (4.35 x 10⁻ ⁶)

22.5

Calculate the potential of a silver electrode in contact with the following: a. a solution that is 0.0150 M in I2 and saturated with AgI. b. a solution that is 0.0040M in CN- and 0.0600M in Ag(CN)2-. c. The solution that results from mixing 25.0mL of 0.0500 KBr with 20.0mL of 0.100 M Ag+ d.

The solution that results from mixing 25.0mL of 0.0500 Ag +with 20.0mL of 0.100 M KBr

Jawab : a)

E = -0.151 – 0.0592 log (0.0150) = -0.043 V

b)

E = -0.31 – 0.0592 log

c)

mmol Br⁻ = 25.0 x 0.0500 = 1.25

(0.0040) ² = -0.10 V 0.0600

mmol Ag+ = 20.0 x 0.100 = 2.00 Kelebihan Br⁻ = 2.00 – 1.25 = 0.75 mmol dalam total volume 45.0 mL [Br⁻] =

0.75 = 0.0167 M 45.0

E = 0.073 – 0.0592 log (0.0167) = 0.178 V 22.6

Calculate the electrode potentials for the following systems: a. Cr2O72- (4.00 x 10-3 M), Cr3+ (2.00 x 10-2 M), H+(0.100 M)|Pt b. UO22++(0.200 M), U4+ (0.100 M), H+ (0.500 M)|Pt Jawab : 0.0592 (2.00 x 10−2 )² log [ ]= 1.20 V 6 (4.00 x 10⁻ ³)(0.100) ¹⁴

a)

E = 1.33 -

b)

E = 0.334 -

0.100 0.0592 log =0.307 V 2 0.200 x (0.600) ⁴

22.7

Calculate the theoretical potential of each of the following cells. Is the cell reaction

spontaneous as written or spontaneous in the opposite direction? a. Pt|Cr3+ (1.00 x 10-4 M), Cr2+ (2.00 x 10 -3 M) || Pb2+ (5.60 x 10 -2 M)|Pb b. Hg|Hg22+ +(2.00 x 10 -2M) || H+ (1.00 x 10 -2M), V3+ (3.00 x 10 -2M), VO (2.00 x 10 -3 M)|Pt c. Pt|Fe3+ (4.00 x 10 -2 M), Fe2+ (3.00 x 10-5 M) || Sn2+ (5.50 x 10 -2 M), Sn4+ (3.50 x 10 -4 M)|Pt Jawab : a)

Ekanan = -0.126 -

0.0592 log ¿-0.159 V 2 2 x 10 ⁻ ³ = -0.485 V 1 x 10⁻ ⁴

Ekiri

= -0.408 – 0.0592 log

Esel

= -0.159 – (-0.485)= 0.326 V

Sel spontan, reaksi oksidasi tertulis di kiri, reduksi di kanan b)

disebelah kanan kita memiliki VO2+ + 2H+ + e- ↔ V3+ + H2O 3.00 x 10⁻ ² = 0.053 V (2.00 x 10−3)(1.00 x 10−2) ²

Ekanan

= 0.359 – 0.0592 log

Ekiri

= 0.788 -

Esel

= 0.053 – 0.738 = -0.685 V

0.0592 1 log = 0.738 V 2 2 x 10⁻ ²

Sel tidak spontan dalam arah yang dipertimbangkan (oksidasi di kiri, reduksi di kanan), tetapi spontan dalam arah yang berlawanan c)

5.50 x 10 ⁻ ² 0.0592 log = 0.089 V 2 ( 3.50 x 10− 4 )

Ekanan

= 0.154 -

Ekiri

= 0.771 – 0.0592 log

Esel

= 0.089 – 0.956 = -0.867 V

3.00 x 10 ⁻ ⁵ = 0.956 V (4.00 x 10−2 )

Sel tidak spontan dalam arah yang dipertimbangkan, melainkan spontan pada arah yang berlawanan (reduksi di kiri, oksidasi di kanan)

22.8

Calculate the theoretical potential of each of the following cells. Is the cell reaction

spontaneous as written or spontaneous in the opposite direction?

a. Bi|BiO+ (0.0400 M), H+ (0.200 M) || I- (0.100 M), AgI (sat'd)|Ag b. Zn|Zn2+(7.50 X 10-4M)||Fe(CN)64- (4.50 x 10-2M),Fe(CN)63- (7.00 x10-2M)| Pt c. Pt2H2 (0.200 atm)|HCI (7.50 X 10-4M), AgCI (sat'd)|Ag Jawab : 1. a)

Ekanan = -0.151 – 0.0592 log 0.100 = -0.092

b)

1 0.0592 log = 0.265 V ( 0.0400 ) (0.200) ² 3

Ekiri

= 0.320 -

Esel

= -0.092 – 0.265 = -0.357 V

Ekanan

= 0.36 – 0.0592 log

Ekiri = -0.763 -

tidak spontan seperti yang tertulis 4.50 x 10 ⁻² −2 = 0.37 V (7.00 x 10 )

1 0.0592 log = 0.265 V 2 (7.00 x 10−⁴ )

Esel = 0.37 – (-0.86) = 1.23 V

spontan seperti yang tertulis (oksidasi di kiri)

c)

Ekanan

= 0.222 – 0.0592 log (7.50 x 10⁻⁴) = 0.407 V 0.200 0.0592 log = -0.164 V −⁴ 2 (7.50 x 10 )²

Ekiri

= 0.000 -

Esel

= 0.407 – (-0.164) = 0.571 V

spontan seperti yang tertulis (oksidasi di kiri)

22.9

Compute Eo for the process Ni(CN)42- + 2e- ⇌ Ni(s) + 4CNGiven that the formation constant for the complex is 1.0 x 1022. Jawab : Ni2+ + 2e– ⇌ Ni(s)

E0 = – 0.250 V

Ni(CN)42– + 2e– ⇌ Ni(s) + 4CN–

E0 = ?

E¿

2+¿/ ¿

¿

= −250−

0.0592 1 log ¿ ¿ ¿ 2

K f =¿ ¿ ¿4 E = −250−

0.0592 log Kf ¿ ¿ ¿ ¿ 2

reaksi kedua : E = −E0 −

0.0592 log Kf ¿ ¿ ¿ ¿ 2

When [CN–]4/[Ni(CN)42–] = 1.00, E = −250− E0 = −250−

0.0592 log Kf 2

0.0592 L og 1 x 1022=−0.90V 2

22.10 The solubility product constant for PbI2 is 7.1 x 10-9 at 25oC. Calculate Eo for the process PbI2 (s) + 2e- ↔ Pb(s) + 2I2 Jawab : Pb2+ + 2e– ⇌ Pb(s)

E0 = – 0.126 V

PbI2 (s) + 2e– ⇌ Pb(s) + 2I–

E0 = ?

E Pb

2+¿/ Pb

¿

= −0.126−

E PbI / Pb = E0 − 2

0.0592 1 log ¿ ¿¿ 2

0.0592 log ¿ ¿¿ ¿ 2

Ksp= ¿ ¿ = 7.1 x 10−9 E

Pb 2+¿/ Pb=−0.126−

0.0592 log¿ ¿¿¿ ¿¿ 2

Dimana : ¿¿ 2

[ 1.00 ] 0.0592 E0 =−0.126− log =−0.367 V 2 7.1 x 10−9 22.11 Calculate the standard potential for the half-reaction BiOCl(s) + 2H+ + 3e– ⇌ Bi(s) + Cl– + H2O Given that Ksp for BiOCI has a value of 8.1 x 10−1 9 Jawab : BiO+ + 2H+ + 3e– ⇌ Bi(s) + H2O

E0 = +0.320 V

BiOCl(s) + 2H+ + 3e– ⇌ Bi(s) + Cl– + H2O

E0 = ?

E BiO

¿

= +0.320−

E BiO

¿

= E0 −

+ ¿/Bi

+ ¿/Bi

[ Bi ] 0.0592 log ¿ ¿ ¿ 3

0.0592 log ¿ ¿¿ 3

Ksp= [BiO+¿ ¿ ¿ = 8.1 x 10−1 9 E

BiO+ ¿/Bi =+ 0.320V −

0.0592 log¿ ¿¿¿ ¿ 3

Dimana : ¿¿ 2

[ 1.00 ] M 0.0592 E0 =+0.320− log 3 8 .1 x 10−19 M E0 =−0.037 V 22.12 Calculate the standard potential for the half-reaction Al ¿ ¿ If the formation constant for the complex is log 1.3 x 1013 Jawab : Al3+ + 3e– ⇌ Al(s) Al ¿ ¿ E Al

3+ ¿/ Al

E0 = – 1.662 V

E0 = ? ¿

= – 1.662 V −

[ Al] 0.0592 log ¿ ¿ ¿ 3

K f =¿ ¿ ¿ E = −1.662 V − 0 E=E −

0.0592 log Kf ¿ ¿ ¿ ¿ ¿ 3

0.0592 log ¿ ¿¿ ¿ ¿ 3

Dimana : ¿¿¿¿ E0 = −1.662 V −

0.0592 log Kf 3

E0 = −1.662 V −

0.0592 log ¿ ¿ ¿ ¿ 3

E0 = −1.662 V −

0.0592 log 1.3 x 1013 3

E0 =−1.92V 22.13 From the standard potentials E0 =−0.336 V

Tl+ + e- ⇌ Tl(s)

TlCl(s) ⇌ Tl(s) + Cl−¿¿ E0 =−0.557 V Calculate the solubility product constant for TlCl Jawab : E = −0.336 V −0.592 log

1 +¿ ¿ [Tl ]

E = −0.336 V −0.592 log

[Cl ]−¿ ¿ Ksp

Dimana : ¿ ¿TlCl = - 0.557 V −0.557 = −0.336 V −0.592 log

1 Ksp

Ksp= 1.85 x 10-4 22.14 From the standard potentials Ag2SeO4(s) + 2e ⇌ 2Ag(s) + SeO42-

E0 =+0.355 V

Ag+ (s) + 2e ⇌ 2Ag(s)

E0 =+0.799 V

Calculate the solubility product constant for TlCl. Jawab : Ag2SeO4(s) + 2e ⇌ 2Ag(s) + SeO42-

E0 =+0.355 V

Ag+ (s) + 2e ⇌ 2Ag(s)

E0 =+0.799 V

E =0.799 V −

[ Ag] 0.0592 log ¿ ¿ ¿ 1

E0 =0.799 V −

0.0592 log ¿ ¿ ¿ 2

0

Dimana : ¿ Ag2SeO4 = 0.355 V 0.355 V

=0.799 V −

0.355 V

=0.799 V −

0.0592 log ¿ ¿ ¿ 2

0.0592 1.00 log 2 [ Ksp ]

0.355 V – 0.799 V = - 0.0296 log

1.00 [ Ksp ]

−0.444 1.00 =log 0.296 [ Ksp ] 15=log 1−log Ksp Log Ksp

= - 15

Ksp

= 1 x 10-15

22.15 Suppose that we wish to produce a current of 0.0750 A in the cell. Pt|V3+ (3.7 x 10-5 M), V2+ (4.48 x 10-1 M) || Br - (0.0850 M). AgBr(sat'd)|Ag As a result of its design, the cell has an internal resistance of 4.87Ω. Calculate the initial potential of the cell. Jawab : E

right

E

= 0.073 – 0.0592 log (0.0850) = 0.136 V

¿=−0.256−0.592 log

0.448 =−0.498V ¿ 3.7 x10 −5

E¿−¿¿= 0.0136 — 0.498=0.634 V Ecell =E¿ −E¿−IR=0.634−0.075 x 4.78=0.268V ¿ 22.16 The cell Pt|V(OH)4+ (2.67 x 10 -4 M), VO2+ (3.42 x l0 -2M), H+ (4.81 x 10-3 M)||Cu2+ (2.50 x 10-2M)|Cu has an internal resistance of 3.81 Ω . What will be the initial potential if a current of 0.0750A is drawn from this cell? Jawab : E

right

= 0.337 – 0.0592log

1 =0.242 V 2.5 x 10−2

E ¿=1.00−0.592log

3.42 x10 −2 ¿¿ (2.67 x10 ¿¿−4 )¿¿¿

E¿−¿¿= 0.242 V — 0.601 V =−0.359 V Ecell =E¿ −E¿−IR =−0.359 V −0.0750 x3.81=−0.644V ¿ 22.17 The resistance of galvanic cell Pt|Fe(CN)64- (4.42 x 10-2M), Fe(CN)63- (8.93 x 10-3 M) || Ag+ (5.75 x 10-1M)|Ag is 3.85 Ω . calculate the initial potential when 0.0442 A is drawn from this cell.

Jawab : Eright = 0.337 – 0.0592 log Eleft = 1,00 – 0,0592 log

1 = 0.242 V 2,50 x 10−2 3,42 x 10−2 2

2,67 x 10−4 x ( 4,81 x 10−3 )

= 0,601 V

Eright – Eleft = 0.242 – 0.601 = –0.359 V Ecell = Eright – Eleft – IR = –0.359 – 0.0750 × 3.81 = –0.644 V Ecell = 0,24 V