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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 0.017 M at 25 °C.

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 0.017 M at 25 °C.

Chemistry
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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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 0.017 M at 25 °C.

2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) → 2 Fe3+(aq) + 2 H2O(ℓ)

Acidic Solution Standard Electrode Potential, E (volts) H2O2(aq) + 2 H*(aq) + 2 e →2 H2O(t) Au*(aq) + e – Au(s) Au3 (aq) + 3 e – Au(s) Br2(t) + 2 e NO: (aq) + 4 H'(aq) + 3 e → Ag(s) +1.77 +1.68 +1.50 →2 Br(aq) +1.08 → NO(g) + 2 H20 +0.96 Ag*(aq) + e Hg2" (aq) + 2 e +0.80 →2 Hg(t) +0.789 Fe (aq) Fe"(aq) + e Cu²"(aq) + 2 e Hg:Clz(s) + 2 e –→ 2 Hg(t) + 2 C(aq) Snt*(aq) + 2e 2 H*(aq) + 2 e Pb?"(aq) + 2 e – Pb(s) Sn2*(aq) + 2 e → Sn(s) Ni?*(aq) + 2 e +0.77 » Cu(s) +0.337 +0.27 Sn2"(aq) +0.15 » H2(g) 0.00 -0.126 -0.14 Ni(s) -0.25 Cd2*(aq) + 2 e – Cd(s) -0.40 Cr*(aq) + e Cr2*(aq) -0.408 Fe2"(aq) + 2 e Zn2"(aq) + 2 e Cr2*( aq) + 2 e A3* (aq) + 3 e Mg2*(aq) + 2 e → Mg(s) Fe(s) -0.44 Zn(s) -0.763 » Cr(s) Al(s) -0.91 -1.66 -2.37 Submit Which of the following is being reduced? O Fe?+ O H,O2 OH Submit
Transcribed Image Text:Acidic Solution Standard Electrode Potential, E (volts) H2O2(aq) + 2 H*(aq) + 2 e →2 H2O(t) Au*(aq) + e – Au(s) Au3 (aq) + 3 e – Au(s) Br2(t) + 2 e NO: (aq) + 4 H'(aq) + 3 e → Ag(s) +1.77 +1.68 +1.50 →2 Br(aq) +1.08 → NO(g) + 2 H20 +0.96 Ag*(aq) + e Hg2" (aq) + 2 e +0.80 →2 Hg(t) +0.789 Fe (aq) Fe"(aq) + e Cu²"(aq) + 2 e Hg:Clz(s) + 2 e –→ 2 Hg(t) + 2 C(aq) Snt*(aq) + 2e 2 H*(aq) + 2 e Pb?"(aq) + 2 e – Pb(s) Sn2*(aq) + 2 e → Sn(s) Ni?*(aq) + 2 e +0.77 » Cu(s) +0.337 +0.27 Sn2"(aq) +0.15 » H2(g) 0.00 -0.126 -0.14 Ni(s) -0.25 Cd2*(aq) + 2 e – Cd(s) -0.40 Cr*(aq) + e Cr2*(aq) -0.408 Fe2"(aq) + 2 e Zn2"(aq) + 2 e Cr2*( aq) + 2 e A3* (aq) + 3 e Mg2*(aq) + 2 e → Mg(s) Fe(s) -0.44 Zn(s) -0.763 » Cr(s) Al(s) -0.91 -1.66 -2.37 Submit Which of the following is being reduced? O Fe?+ O H,O2 OH Submit
iing Whi H0, ald Wh in Q-F' H,OH 1- Whtie H0, -LV The Nodudim ed dk i
Transcribed Image Text:iing Whi H0, ald Wh in Q-F' H,OH 1- Whtie H0, -LV The Nodudim ed dk i
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