hello i need help with part a and b.1a) Using tabulated standard reduction potentials from the table i attached below, calculate the standard cell potential, E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:IO3−(aq) + Fe2+(aq) ⇌ Fe3+(aq) + I2(s)E°cell =1b) A galvanic cell based on the above reaction is constructed according to the generic sketch of a galvanic cell shown in the picture i attached.Choose the letter from the picture which best describes the following:-Solution containing IO3−, I2 and source of H+ -Pt cathode Ee-FHGDACB Standard Reduction Potentials at 25°C (298 K) for Many Common Half-ReactionsHalf-Reaction8° (V)Half-ReactionF₂ +2e7 →2F-O₂ + 2H₂O + 4e¯Cu²+ + 2e-Ag2+ + eCo³+ + eAg+Co²+H₂O₂ + 2H+ + 2e →→→ 2H₂OCe4+ + eCe ³++2ePbO₂ + 4H+ + SO4²MnO4 + 4H+ + 3e-IO4 + 2H+ + 2e7MnO4 + 8H+ + Se7Au³+ + 3eAuPbO₂ + 4H+ + 2eCl₂ +2e72C1-Cr₂O7²- + 14H+ + 6e7O₂ + 4H+ + 4e7MnO₂ + 4H+ + 2e¯IO3 + 6H+ + SeBr₂ +2e →→→ 2BrAg+ + e-Hg₂+ + 2e-Fe³+ + e- →→ Fe²+ClO₂-→Hg₂2+Ag2HgMnO₂ + 2H₂O103 + H₂OCu2H₂OPbSO4 + 2H₂OMn²+ + 4H₂OPb²+ + 2H₂OVO₂+ + 2H+ + e-AuCl4 + 3e Au + 4Cl-NO3 + 4H+ + 3e¯ClO₂ + e-2Hg2+ + 2e72Cr³+ + 7H₂OMn²+ + 2H₂O1₂ + 3H₂OVO²+ + H₂OO₂ + 2H+ + 2e7 →→→→→ H₂O₂MnO4 + eMnO4²-1₂ +2e7 -21-CuteNO + 2H₂O2.871.991.821.781.701.691.681.601.511.501.461.361.331.231.211.201.091.000.990.960.9540.910.800.800.770.680.560.540.52Fe³+ + 3e-Pb²+ + 2e7Sn²+ + 2e-Ni²+ + 2e7Hg₂Cl₂ +2e7AgCl + e →Ag + Cl-SO4 + 4H+ + 2e7Cu²+ + e2H+ + 2e-PbSO4 + 2e-Cd²+ + 2e7Fe²+ + 2e7Cr³+ + eCr³+ + 3e7Zn²+ + 2e72H₂O + 2e7Mn²+ + 2eAl³+ + 3e-H₂ +2e7Mg2+ + 2e-La³+ + 3e7Na+ + eCa2+ + 2e7Ba2+2e7K+ + eLiteCuCu+H₂FePbSnNiK2Hg + 2C1-LiCdAl2H-FeCr²+CrZnH₂ + 2OH-MnNaPb + SO4²-Mg40H-LaCaH₂SO3 + H₂OBa8° (V)0.400.340.270.220.200.160.00-0.036-0.13-0.14-0.23-0.35-0.40-0.44-0.50-0.73-0.76-0.83-1.18-1.66-2.23-2.37-2.37-2.71-2.76-2.90-2.92-3.05

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Answered: E e- F H G D A C B

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 150CP: Given the following two standard reduction potentials, solve for the standard reduction potential of...

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hello i need help with part a and b.

1a) Using tabulated standard reduction potentials from the table i attached below, calculate the standard cell potential, E°_cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

IO₃⁻(aq) + Fe²⁺(aq) ⇌ Fe³⁺(aq) + I₂(s)

E°_cell =

1b) A galvanic cell based on the above reaction is constructed according to the generic sketch of a galvanic cell shown in the picture i attached.

Choose the letter from the picture which best describes the following:
-Solution containing IO₃⁻, I₂ and source of H⁺

-Pt cathode

Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions
Half-Reaction
8° (V)
Half-Reaction
F₂ +2e7 →2F-
O₂ + 2H₂O + 4e¯
Cu²+ + 2e-
Ag2+ + e
Co³+ + e
Ag+
Co²+
H₂O₂ + 2H+ + 2e →→→ 2H₂O
Ce4+ + e
Ce ³+
+2e
PbO₂ + 4H+ + SO4²
MnO4 + 4H+ + 3e-
IO4 + 2H+ + 2e7
MnO4 + 8H+ + Se7
Au³+ + 3e
Au
PbO₂ + 4H+ + 2e
Cl₂ +2e7
2C1-
Cr₂O7²- + 14H+ + 6e7
O₂ + 4H+ + 4e7
MnO₂ + 4H+ + 2e¯
IO3 + 6H+ + Se
Br₂ +2e →→→ 2Br
Ag+ + e-
Hg₂+ + 2e-
Fe³+ + e- →→ Fe²+
ClO₂-
→Hg₂2+
Ag
2Hg
MnO₂ + 2H₂O
103 + H₂O
Cu
2H₂O
PbSO4 + 2H₂O
Mn²+ + 4H₂O
Pb²+ + 2H₂O
VO₂+ + 2H+ + e-
AuCl4 + 3e Au + 4Cl-
NO3 + 4H+ + 3e¯
ClO₂ + e-
2Hg2+ + 2e7
2Cr³+ + 7H₂O
Mn²+ + 2H₂O
1₂ + 3H₂O
VO²+ + H₂O
O₂ + 2H+ + 2e7 →→→→→ H₂O₂
MnO4 + e
MnO4²-
1₂ +2e7 -21-
Cute
NO + 2H₂O
2.87
1.99
1.82
1.78
1.70
1.69
1.68
1.60
1.51
1.50
1.46
1.36
1.33
1.23
1.21
1.20
1.09
1.00
0.99
0.96
0.954
0.91
0.80
0.80
0.77
0.68
0.56
0.54
0.52
Fe³+ + 3e-
Pb²+ + 2e7
Sn²+ + 2e-
Ni²+ + 2e7
Hg₂Cl₂ +2e7
AgCl + e →Ag + Cl-
SO4 + 4H+ + 2e7
Cu²+ + e
2H+ + 2e-
PbSO4 + 2e-
Cd²+ + 2e7
Fe²+ + 2e7
Cr³+ + e
Cr³+ + 3e7
Zn²+ + 2e7
2H₂O + 2e7
Mn²+ + 2e
Al³+ + 3e-
H₂ +2e7
Mg2+ + 2e-
La³+ + 3e7
Na+ + e
Ca2+ + 2e7
Ba2+2e7
K+ + e
Lite
Cu
Cu+
H₂
Fe
Pb
Sn
Ni
K
2Hg + 2C1-
Li
Cd
Al
2H-
Fe
Cr²+
Cr
Zn
H₂ + 2OH-
Mn
Na
Pb + SO4²-
Mg
40H-
La
Ca
H₂SO3 + H₂O
Ba
8° (V)
0.40
0.34
0.27
0.22
0.20
0.16
0.00
-0.036
-0.13
-0.14
-0.23
-0.35
-0.40
-0.44
-0.50
-0.73
-0.76
-0.83
-1.18
-1.66
-2.23
-2.37
-2.37
-2.71
-2.76
-2.90
-2.92
-3.05

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