rtial, EO (7.7) O p 2H 4OH ( A voltaic cell based on the half- reactions: In (aq) → In*³(aq) + 2e¯ Br2() +2e- 2Br¯(aq) The standard emf for this cell is +1.46 V. Using the data below, calculate, E red for the reduction of In*3 to In*1. Br2() + 2e - 2Br (ag) (cathode) E red(cathode) = E Br(aq) = +1.06 V %3D %3D (See Table below) Choices: (A) +0.40 V (B) -0.40 V (C) + 2.50 V (D) -2.50 V Answer: Ered for the reduction of In+3 to In+1 = (in %3D volts) STANDARD REDUCTION POTENTIALS at 25°C; Reduction Half Reaction F(g) + 2e 2 (ag HO (ag) + 2H" (ag) + 2H0 PO)- (ag) - sotag) + 2e PosOs) - HO MaO (ag)+ Hr (a)+ SeMa (ag)+ 4H0 Au (a) - e Au (s) + 2e 20 (a CrO (ag)- 14H" (ag+ 6e +2C (ag)+ THO MaO: (s) 4H (ag de Ma (ag) 2H:0 O: 0 4 ( de 1o Br () + 2 2Br (ag) NO, (ag) + 4r (ag) + e+ NO (g) + 210 Standard Potential, E(V) +287 +1.77 1.70 +151 +1.50 +136 +133 1.23 +123 +1.06 +0.96 +0.92 +0.85 0.30 +0.77 068 +0.59 Ag a e Ag is) Feag) Te (a O (g)- 21 (ag) +2e 1O, (ag I MaO (ag) + 210 - 3e MeO; (s) - 401 (ag) I) + 2e 21 (ag) O (- 2HO (ag) - de+ 40H" (ag) Cu" (ag)+ 2e Cu ) AgCI (s)+e Ag (s)- CI (ag) só (ag) + 4 ()+ 2e so ( + 2H:O +054 +040 +0.34 +0.22 +0.20 +0.15 0.13 0.00 * Se () 2e Sa" (a) 4 21' () 2r H: ) a) 2e Se" (ag)+ 2e Sa (s) * N (g)- 2e Ni (s) * PSo, )+ 2e- P() so (g C (ag) + C (a) Fe (aq)+ 2 Fe (s) C (ag)- e Cr() Za (ag) + 2e Za (s) 2HO + 2e H ig)+ 20H () Ma (ag) + 2e Ma (s) Al (a e Al () Mg" (ag)+ 2e Mgis) Na (ag)+e→ Na (s) Ca(ag)+ K (a)e K L(ag) +e → Li) For all hall ceactices the coacentrations is IM for dissolved species and the pressure is I at for gases. These are the standaed-state valses -0.14 0.25 031 0.41 044 -0.74 -0.76 -0.83 -L18 -1.66 -2.37 -2.71 2e Ca (s) -2.87 -2.93 -3.05

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ISBN:9781305957404
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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(7.7)
A voltaic cell based on the half-
reactions:
In (aq)
– In*3(ag) + 2e™
(aq)
Br2() +2e¯→ 2Br¯(ag)
The standard emf for this cell is +1.46 V.
Using the data below, calculate, E'red for
the reduction of In*3 to In*1.
Br2() + 2e¯ → 2B1¯(ag) (cathode)
E red(cathode) = E
%3D
Br(aq) = +1.06 V
(See Table below)
Choices:
(A) +0.40 V
(B) -0.40 V
(C) + 2.50 V
(D) -2.50 V
Answer: E'red for the reduction of
In+3 =
to In*1
(in
volts)
STANDARD REDUCTION POTENTIALS at 25°C;
Reduction Half Reaction
F: (g) + 2e → 2F (aq)
HO, (aq) + 2H" (aq) + 2e 2H;0
PbO(3) +4H" (ag) + So (ag) + 2e PbSO(s) + HO
MnO. (aq) + 8H" (aq)+ Se+ Ma" (ag) + 4H,o
Au" (aq) + 3e Au (s)
Ch (g) + 2e 2C (ag)
CrO (aq) + 14H" (aq) + 6e →2Cr" (aq) + TH:O
MnO: (3) i 4H" (aq) 2e Ma (aq) 2H:0
O: (g) + 4H" (aq) + 4e + 2H:0
Br: () +2e + 2Br (ag)
NO, (aq) + 4I" (aq) + 3e + NO (g) + 2H:0
2Hg (aq) + 2e Hg" (aq)
Hgr" (aq) + 2e→ 2Hg ()
Ag (aq) +e Ag (s)
Fe (aq) + e Fe (aq)
O: (g) - 21 (aq) + 2e HO: (ag)
E MnO" (aq) + 2H0 + 3e MnO: (3) + 40H (aq)
E L (s) + 20 → 21 (aq)
O2 (g) + 2H:0 (aq) + 4e + 40H" (aq)
I
Standard Potential, E°(V)
+2.87
+1.77
+1.70
+1.51
+1.50
+1.36
+1.33
11.23
+1.23
+1.06
+0.96
+0.92
+0.85
+0.80
7
+0.77
+0.68
+0.59
+0 54
+0.40
Cu" (aq) + 2e" Cu (s)
ABCI (s) + e Ag (s) + CI (aq)
SO, (aq) + 4H" (aq) + 2e + SO: (g) + 2H;0
* Cu" (aq) + e Cu (ag)
4 Sat" (ag) + 2e Sa" (aq)
2H' (aq) + 2e→ H: (4)
I Pb (aq) + 2e+ Pb (s)
Sn" (aq) + 2e Sn (s)
* Ni (aq) + 2e + Ni (s)
* PBSO, (s) + 2e Pb (s) + So (aq)
* Cr" (aq) + e C (ag)
i Fe (aq) + 2e+ Fe (s)
Cr (aq) + 3e → Cr (s)
Zn- (aq) + 2e→ Zn (s)
2H:0 + 2e Hi (g) + 20H (aq)
Mn* (aq) + 2e → Mn (s)
Al (aq) + 3e → Al (s)
Mg (aq) + 2e - Mg (s)
Na" (aq) + e- Na (s)
Ca" (aq) + 2e → Ca (s)
K* (aq) + e K (3)
Li" (aq) + e + Li (s)
For all bal reactions the concentrations is IM for dissolved species and the pressure is
1 atm for gases. These are the standard-state values.
+034
+0.22
+0.20
+0.15
+0.13
0.00
-0.13
-0.14
-0.25
-0.31
-0.41
-0.44
-0.74
-0.76
0.83
-1.18
-1.66
-2.37
-2.71
-2.87
-2.93
-3.05
uade Buonpaa se gäuans SurseaDuI
Transcribed Image Text:(7.7) A voltaic cell based on the half- reactions: In (aq) – In*3(ag) + 2e™ (aq) Br2() +2e¯→ 2Br¯(ag) The standard emf for this cell is +1.46 V. Using the data below, calculate, E'red for the reduction of In*3 to In*1. Br2() + 2e¯ → 2B1¯(ag) (cathode) E red(cathode) = E %3D Br(aq) = +1.06 V (See Table below) Choices: (A) +0.40 V (B) -0.40 V (C) + 2.50 V (D) -2.50 V Answer: E'red for the reduction of In+3 = to In*1 (in volts) STANDARD REDUCTION POTENTIALS at 25°C; Reduction Half Reaction F: (g) + 2e → 2F (aq) HO, (aq) + 2H" (aq) + 2e 2H;0 PbO(3) +4H" (ag) + So (ag) + 2e PbSO(s) + HO MnO. (aq) + 8H" (aq)+ Se+ Ma" (ag) + 4H,o Au" (aq) + 3e Au (s) Ch (g) + 2e 2C (ag) CrO (aq) + 14H" (aq) + 6e →2Cr" (aq) + TH:O MnO: (3) i 4H" (aq) 2e Ma (aq) 2H:0 O: (g) + 4H" (aq) + 4e + 2H:0 Br: () +2e + 2Br (ag) NO, (aq) + 4I" (aq) + 3e + NO (g) + 2H:0 2Hg (aq) + 2e Hg" (aq) Hgr" (aq) + 2e→ 2Hg () Ag (aq) +e Ag (s) Fe (aq) + e Fe (aq) O: (g) - 21 (aq) + 2e HO: (ag) E MnO" (aq) + 2H0 + 3e MnO: (3) + 40H (aq) E L (s) + 20 → 21 (aq) O2 (g) + 2H:0 (aq) + 4e + 40H" (aq) I Standard Potential, E°(V) +2.87 +1.77 +1.70 +1.51 +1.50 +1.36 +1.33 11.23 +1.23 +1.06 +0.96 +0.92 +0.85 +0.80 7 +0.77 +0.68 +0.59 +0 54 +0.40 Cu" (aq) + 2e" Cu (s) ABCI (s) + e Ag (s) + CI (aq) SO, (aq) + 4H" (aq) + 2e + SO: (g) + 2H;0 * Cu" (aq) + e Cu (ag) 4 Sat" (ag) + 2e Sa" (aq) 2H' (aq) + 2e→ H: (4) I Pb (aq) + 2e+ Pb (s) Sn" (aq) + 2e Sn (s) * Ni (aq) + 2e + Ni (s) * PBSO, (s) + 2e Pb (s) + So (aq) * Cr" (aq) + e C (ag) i Fe (aq) + 2e+ Fe (s) Cr (aq) + 3e → Cr (s) Zn- (aq) + 2e→ Zn (s) 2H:0 + 2e Hi (g) + 20H (aq) Mn* (aq) + 2e → Mn (s) Al (aq) + 3e → Al (s) Mg (aq) + 2e - Mg (s) Na" (aq) + e- Na (s) Ca" (aq) + 2e → Ca (s) K* (aq) + e K (3) Li" (aq) + e + Li (s) For all bal reactions the concentrations is IM for dissolved species and the pressure is 1 atm for gases. These are the standard-state values. +034 +0.22 +0.20 +0.15 +0.13 0.00 -0.13 -0.14 -0.25 -0.31 -0.41 -0.44 -0.74 -0.76 0.83 -1.18 -1.66 -2.37 -2.71 -2.87 -2.93 -3.05 uade Buonpaa se gäuans SurseaDuI
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