Voltaic Cells Under Non-Standard Conditions Step 1-2 Identify the reduction half-reaction. E° for this reaction is E°cathode- Step 3-4 Identify the oxidation half-reaction. E° for this reaction is E°anode- Step 5 Determine the number of electrons transferred in the process. Step 6 Calculate E° cell- E°cell = E°cathode - E°anode Step 7 Identify the form of the reaction quotient, Q. This is of the same form as the equilibrium constant expression for the system being studied. Step 8 Calculate the numerical value of Q. Step 9 Use the Nernst equation to calculate the cell potential. Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.025 M at 25 °C. 2 Fe (aq) + H,O2(aq) + 2 H¨(aq) → 2 Fe¯(aq) + 2 H,O(t) Acidic Solution Standard Electrode Potential, E(volts)

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...
icon
Related questions
Question
TUTOR
Voltaic Cells Under Non-Standard Conditions
Step 1-2
Identify the reduction half-reaction. E° for this reaction is E°cathode:
Step 3-4
Identify the oxidation half-reaction. E° for this reaction is E°,
anode-
Step 5
Determine the number of electrons transferred in the process.
Step 6
Calculate E°cell-
E°cell = E°cathode - E°anode
Step 7
Identify the form of the reaction quotient, Q. This is of the same form as the equilibrium constant expression for the
system being studied.
Step 8
Calculate the numerical value of Q-
Step 9
Use the Nernst equation to calculate the cell potential.
Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.025 M at 25
°C.
2 Fe (aq) + H2O2(aq) + 2 H*(aq) – 2 Fe"(aq) + 2 H20(E)
Acidic Solution
Standard Electrode Potential, E'(volts)
H2O2(aq) + 2 H'(aq) + 2 e 2 H20(t)
+1.77
Au*(aq) + e - Au(s)
Au"(ag) + 3 e →Au(s)
Br2(e) + 2 e -→2 Br(ag)
NO3 (aq) + 4 H'(aq) + 3 e → NO(g) + 2 H20
+1.68
+1.50
+1.08
+0.96
+0.80
Ag*(aq) + e – Ag(s)
Hg22"(aq) + 2 e 2 Hg(e)
Fe"(ag) + e- → Fe2"(ag)
→ Cu(s)
+0.789
+0.77
Cu2"(ag) + 2 e
+0.337
Hg2Clz(s) + 2 e
2 Hg(t) + 2 C(aq)
+0.27
Sn+*(aq) + 2e → Sn2*(aq)
+0.15
→ Hz(g)
→ Pb(s)
→ Sn(s)
Ni(s)
2 H*(aq) + 2 e
0.00
Pb"(aq) + 2 e
Sn2*(aq) + 2 e
Ni2"(aq) + 2 e
Cd*(aq) + 2 e » Cd(s)
Cr**(aq) + e – Cr*(aq)
Fe2*(aq) + 2 e Fe(s)
Zn2*(aq) + 2 e → Zn(s)
Cr2*( aq) + 2 e
Al3+ (ag) + 3 e
-0.126
-0.14
-0.25
-0.40
-0.408
-0.44
-0.763
→ Cr(s)
> Al(s)
-0.91
-1.66
Mg"(aq) + 2 e
Mg(s)
-2.37
V
Submit
Transcribed Image Text:TUTOR Voltaic Cells Under Non-Standard Conditions Step 1-2 Identify the reduction half-reaction. E° for this reaction is E°cathode: Step 3-4 Identify the oxidation half-reaction. E° for this reaction is E°, anode- Step 5 Determine the number of electrons transferred in the process. Step 6 Calculate E°cell- E°cell = E°cathode - E°anode Step 7 Identify the form of the reaction quotient, Q. This is of the same form as the equilibrium constant expression for the system being studied. Step 8 Calculate the numerical value of Q- Step 9 Use the Nernst equation to calculate the cell potential. Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.025 M at 25 °C. 2 Fe (aq) + H2O2(aq) + 2 H*(aq) – 2 Fe"(aq) + 2 H20(E) Acidic Solution Standard Electrode Potential, E'(volts) H2O2(aq) + 2 H'(aq) + 2 e 2 H20(t) +1.77 Au*(aq) + e - Au(s) Au"(ag) + 3 e →Au(s) Br2(e) + 2 e -→2 Br(ag) NO3 (aq) + 4 H'(aq) + 3 e → NO(g) + 2 H20 +1.68 +1.50 +1.08 +0.96 +0.80 Ag*(aq) + e – Ag(s) Hg22"(aq) + 2 e 2 Hg(e) Fe"(ag) + e- → Fe2"(ag) → Cu(s) +0.789 +0.77 Cu2"(ag) + 2 e +0.337 Hg2Clz(s) + 2 e 2 Hg(t) + 2 C(aq) +0.27 Sn+*(aq) + 2e → Sn2*(aq) +0.15 → Hz(g) → Pb(s) → Sn(s) Ni(s) 2 H*(aq) + 2 e 0.00 Pb"(aq) + 2 e Sn2*(aq) + 2 e Ni2"(aq) + 2 e Cd*(aq) + 2 e » Cd(s) Cr**(aq) + e – Cr*(aq) Fe2*(aq) + 2 e Fe(s) Zn2*(aq) + 2 e → Zn(s) Cr2*( aq) + 2 e Al3+ (ag) + 3 e -0.126 -0.14 -0.25 -0.40 -0.408 -0.44 -0.763 → Cr(s) > Al(s) -0.91 -1.66 Mg"(aq) + 2 e Mg(s) -2.37 V Submit
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY