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What is the value of n, the number of electrons transferred during the reaction? H2O2 reduced at cathode Fe2* oxidized at anode Submit What is the value of E° for the reaction, in volts? H,O2 reduced at cathode Fe2* oxidized at anode 2 e transferred

What is the value of n, the number of electrons transferred during the reaction? H2O2 reduced at cathode Fe2* oxidized at anode Submit What is the value of E° for the reaction, in volts? H,O2 reduced at cathode Fe2* oxidized at anode 2 e transferred

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What is the value of n, the number of electrons transferred during the reaction? H,O2 reduced at cathode Fe2* oxidized at anode Submit What is the value of E° for the reaction, in volts? H2O2 reduced at cathode Fe2* oxidized at anode 2 e transferred V Submit
Transcribed Image Text:What is the value of n, the number of electrons transferred during the reaction? H,O2 reduced at cathode Fe2* oxidized at anode Submit What is the value of E° for the reaction, in volts? H2O2 reduced at cathode Fe2* oxidized at anode 2 e transferred V Submit
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) Aus (aq) + 3 e →Au(s) Br2(l) + 2 e - 2 Br(ag) NO: (aq) + 4 H*(aq) + 3 e NO(g) + 2 H20 Ag*(aq) + e - Ag(s) Hg22"(aq) + 2 e →2 Hg(?) +1.68 +1.50 +1.08 +0.96 +0.80 +0.789 Fe"(ag) + e- → Fe2"(ag) +0.77 Cu2*(aq) + 2 e → Cu(s) +0.337 Hg2Clz(s) + 2 e 2 Hg(t) + 2 Cr(ag) +0.27 Sn**(aq) + 2e 2 H*(aq) + 2 e Pb2*(aq) + 2 e Sn2*(aq) + 2 e Sn2"(aq) H2(g) +0.15 0.00 -0.126 Pb(s) → Sn(s) Ni2*(aq) + 2 e → Ni(s) → Cd(s) Cr*(aq) + e - Cr*(aq) Fe2*(aq) + 2 e Fe(s) → Zn(s) Cr(s) Al3+ (aq) + 3 e Al(s) -0.14 -0.25 Cd2*(ag) + 2 e -0.40 -0.408 -0.44 Zn2"(aq) + 2 e Cr**( ag) + 2 e -0.763 -0.91 -1.66 Mg2*(aq) + 2 e → Mg(s) -2.37
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) Aus (aq) + 3 e →Au(s) Br2(l) + 2 e - 2 Br(ag) NO: (aq) + 4 H*(aq) + 3 e NO(g) + 2 H20 Ag*(aq) + e - Ag(s) Hg22"(aq) + 2 e →2 Hg(?) +1.68 +1.50 +1.08 +0.96 +0.80 +0.789 Fe"(ag) + e- → Fe2"(ag) +0.77 Cu2*(aq) + 2 e → Cu(s) +0.337 Hg2Clz(s) + 2 e 2 Hg(t) + 2 Cr(ag) +0.27 Sn**(aq) + 2e 2 H*(aq) + 2 e Pb2*(aq) + 2 e Sn2*(aq) + 2 e Sn2"(aq) H2(g) +0.15 0.00 -0.126 Pb(s) → Sn(s) Ni2*(aq) + 2 e → Ni(s) → Cd(s) Cr*(aq) + e - Cr*(aq) Fe2*(aq) + 2 e Fe(s) → Zn(s) Cr(s) Al3+ (aq) + 3 e Al(s) -0.14 -0.25 Cd2*(ag) + 2 e -0.40 -0.408 -0.44 Zn2"(aq) + 2 e Cr**( ag) + 2 e -0.763 -0.91 -1.66 Mg2*(aq) + 2 e → Mg(s) -2.37
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