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Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(l) -

Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(l) -

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.148QP: Consider the following cell reaction at 25C. 2Cr(s)+3Fe2+(aq)2Cr3+(aq)+3Fe(s) Calculate the standard...
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Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(1) Acidic Solution Standard Electrode Potential, E (volts) H₂O₂(aq) + 2 H*(aq) + 2 e →→ 2 H₂O(l) +1.77 Aut(aq) + e — Au(s) +1.68 +1.50 Au"(aq) + 3e — Au(s) Brz(t) + 2 e 2 Br(aq) +1.08 NO3(aq) + 4 H (aq) + 3 e NO(g) + 2 H₂O +0.96 Ag*(aq) + e → → Ag(s) +0.80 Hgz²(aq) + 2 e 2 Hg(l) +0.789 Fe³(aq) + e-→→→→→ Fe²(aq) +0.77 Cu²(aq) + 2 e - Cu(s) +0.337 HgzClz(s) + 2 e2 Hg(t) + 2 Cl(aq) +0.27 Sn(aq) + 2e →→→ Sn²+ (aq) +0.15 0.00 2 H*(aq) + 2 e →→→H₂(g) Pb²(aq) + 2e →→→→ Pb(s) -0.126 Sn²(aq) + 2 e-> Sn(s) -0.14 Ni²+ (aq) + 2 e → Ni(s) -0.25 -0.40 Cd²+ (aq) + 2 e → Cd(s) Cr³(aq) + e Cr²+ (aq) -0.408 Fe²(aq) + 2 e→→→→> Fe(s) -0.44 Zn² (aq) + 2 e→→→→→→ Zn(s) -0.763 Cr²(aq) + 2 e Cr(s) -0.91 Al³+ (aq) + 3 e-> Al(s) -1.66 Mg²+ (aq) + 2 e →→→Mg(s) -2.37 V
Transcribed Image Text:Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(1) Acidic Solution Standard Electrode Potential, E (volts) H₂O₂(aq) + 2 H*(aq) + 2 e →→ 2 H₂O(l) +1.77 Aut(aq) + e — Au(s) +1.68 +1.50 Au"(aq) + 3e — Au(s) Brz(t) + 2 e 2 Br(aq) +1.08 NO3(aq) + 4 H (aq) + 3 e NO(g) + 2 H₂O +0.96 Ag*(aq) + e → → Ag(s) +0.80 Hgz²(aq) + 2 e 2 Hg(l) +0.789 Fe³(aq) + e-→→→→→ Fe²(aq) +0.77 Cu²(aq) + 2 e - Cu(s) +0.337 HgzClz(s) + 2 e2 Hg(t) + 2 Cl(aq) +0.27 Sn(aq) + 2e →→→ Sn²+ (aq) +0.15 0.00 2 H*(aq) + 2 e →→→H₂(g) Pb²(aq) + 2e →→→→ Pb(s) -0.126 Sn²(aq) + 2 e-> Sn(s) -0.14 Ni²+ (aq) + 2 e → Ni(s) -0.25 -0.40 Cd²+ (aq) + 2 e → Cd(s) Cr³(aq) + e Cr²+ (aq) -0.408 Fe²(aq) + 2 e→→→→> Fe(s) -0.44 Zn² (aq) + 2 e→→→→→→ Zn(s) -0.763 Cr²(aq) + 2 e Cr(s) -0.91 Al³+ (aq) + 3 e-> Al(s) -1.66 Mg²+ (aq) + 2 e →→→Mg(s) -2.37 V
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