Use the Nernst equation to calculate the actual cell potential ofa galvanic cell under the following non-standard conditions:½ cell 1: Pt electrode1.81 M Fe2+0.42 M Fe3+½ cell 2: Zn electrode2.89 M Zn2+Temperature: 30.09 °CF = 96,485.34 C/ mol e%3DR = 8.314462J/ K-molUse this Table of Standard Reduction PotentialsIMPORTANT: When entering your answer:• Enter the number only (no units)• Do not leave any spaces• Include the sign if your answer is negative (not necessary forpositive values)• Use a leading zero before the decimal when necessaryReport your number to 2 decimal places (regardless of thesignificant figures)Correctly round your answer to the 2nd decimal place (thereis no margin of allowed error in this question, so always useun-rounded numbers in your calculations), If the last digits F2 + 2 e → 2 FS,O, + 2 e → 2 SO,-Co + e → Co*MnO, + 8 H + 5 e → Mn²* + 4 H,OAu + 3 e Au (s)2.872.101.821.511.50C2+ 2 e → 2 CIO2 + 4 H + 4 e → 2 H½O1.361.23Br2 (1) + 2 e → 2 Br2 Hg** + 2 e → Hg2+ 2 e → Hg)1.070.92HgAgHg2* + 2 e → 2 Hg (1)Fe* + e → Fe*0.85e → Ag(9)0.800.790.77I2(6) + 2 e → 2 FCu* + e → Cu(s)Cu+ + 2 e → Cus)0.530.520.34Cu* + e → Cu"0.15Sn** + 2 e → Sn²*S) + 2 H' + 2 e → H,S (2)2 H* + 2 e → H2 (g)Fe + 3 e Fe)Pb0.150.140.00-0.04+ 2 e → Pb (s)Sn2+ + 2 e-0.13Sn.0 14

The reduction half cell reaction is, The oxidation half cell reaction is, Zn2+(aq) + 2e-⇌Zn(s)…

Use the Nernst equation to calculate the actual cell potential of a galvanic cell under the following non-standard conditions: 2 cell 1: Pt electrode 1.81 M Fe2+ 0.42 M Fe3+ 2 cell 2: Zn electrode 2.89 M Zn2+ Temperature: 30.09 °C F = 96,485.34 C/ mol e R = 8.314462J/ K-mol

Use the Nernst equation to calculate the actual cell potential of a galvanic cell under the following non-standard conditions: 2 cell 1: Pt electrode 1.81 M Fe2+ 0.42 M Fe3+ 2 cell 2: Zn electrode 2.89 M Zn2+ Temperature: 30.09 °C F = 96,485.34 C/ mol e R = 8.314462J/ K-mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

Use the Nernst equation to calculate the actual cell potential of
a galvanic cell under the following non-standard conditions:
½ cell 1: Pt electrode
1.81 M Fe2+
0.42 M Fe3+
½ cell 2: Zn electrode
2.89 M Zn2+
Temperature: 30.09 °C
F = 96,485.34 C/ mol e
%3D
R = 8.314462J/ K-mol
Use this Table of Standard Reduction Potentials
IMPORTANT: When entering your answer:
• Enter the number only (no units)
• Do not leave any spaces
• Include the sign if your answer is negative (not necessary for
positive values)
• Use a leading zero before the decimal when necessary
Report your number to 2 decimal places (regardless of the
significant figures)
Correctly round your answer to the 2nd decimal place (there
is no margin of allowed error in this question, so always use
un-rounded numbers in your calculations), If the last digits

F2 + 2 e → 2 F
S,O, + 2 e → 2 SO,-
Co + e → Co*
MnO, + 8 H + 5 e → Mn²* + 4 H,O
Au + 3 e Au (s)
2.87
2.10
1.82
1.51
1.50
C2+ 2 e → 2 CI
O2 + 4 H + 4 e → 2 H½O
1.36
1.23
Br2 (1) + 2 e → 2 Br
2 Hg** + 2 e → Hg2
+ 2 e → Hg)
1.07
0.92
Hg
Ag
Hg2* + 2 e → 2 Hg (1)
Fe* + e → Fe*
0.85
e → Ag(9)
0.80
0.79
0.77
I2(6) + 2 e → 2 F
Cu* + e → Cu(s)
Cu+ + 2 e → Cus)
0.53
0.52
0.34
Cu* + e → Cu"
0.15
Sn** + 2 e → Sn²*
S) + 2 H' + 2 e → H,S (2)
2 H* + 2 e → H2 (g)
Fe + 3 e Fe)
Pb
0.15
0.14
0.00
-0.04
+ 2 e → Pb (s)
Sn2+ + 2 e
-0.13
Sn.
0 14

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