888 Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.011 M at 25 °C.2 Fe2"(aq) + H2O2(aq) + 2 H* (aq) → 2 Fe"(aq) + 2 H20(£)Acidic SolutionStandard Electrode Potential, E'(volts)H2O2(aq) + 2 H*(aq) + 2 e -Au*(aq) + e → Au(s)Au3 (aq) + 3 e – Au(s)2 H2O(t)+1.77+1.68+1.50Br2(e) + 2 e –2 Br(aq)+1.08NO: (aq) + 4 H'(aq) + 3 eAg*(aq) + eHg22"(aq) + 2 e –→2 Hg(t)NO(g) + 2 H20+0.96→ Ag(s)+0.80+0.789Fe"(aq) + e- → Fe2"(aq)Cu²"(aq) + 2 e+0.77Cu(s)+0.337Hg:Cl2(s) + 2 eSn*(aq) + 2e →2 H*(aq) + 2 e H2(g)Pb2"(aq) + 2 e→2 Hg(t) + 2 CI(aq)+0.27Sn2*(aq)+0.150.00» Pb(s)Sn(s)Ni2"(aq) + 2 e → Ni(s)> Cd(s)Cr**(aq)-0.126Sn2*(aq) + 2 e-0.14-0.25Cd2"(aq) + 2 e -Cr*(aq) + e →-0.40-0.408Fe2"(aq) + 2 e– Fe(s)-0.44>Zn²"(aq) + 2 e – Zn(s)-0.763Cr*( aq) + 2 e» Cr(s)-0.91A* (aq) + 3 e> Al(s)-1.66Mg"(aq) + 2 e – Mg(s)-2.37VSubmitShow Approach Show Tutor Steps

Given reaction: 2 Fe2+ (aq) + H2O2 (aq) + 2H+ (aq) → 2 Fe3+ (aq) + 2 H2O (l) It is given that all…

Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.011 M at 25 °C. 2 Fe2"(aq) + H2O2(aq) + 2 H¨(aq) → 2 Fe (aq) + 2 H,O(t) Acidic Solution Standard Electrode Potential, E(volts) H2O2{aq) + 2 H'(aq) + 2 e - 2 H2O(t) +1.77 Au*(aq) + e →Au(s) +1.68 Au (aq) + 3 e-→ Au(s) Brz(8) + 2 e –2 Br(aq) NO: (aq) + 4 H*(aq) + 3 e → NO(g) + 2 H20 Ag*(aq) + e +1.50 +1.08 +0.96 →Ag(s) +0.80 Hg2"(aq) + 2 e –→2 Hg(?) +0.789 Fe (aq) + e-→ Fe²"(aq) Cu²"(aq) + 2 e – Hg:Clz(s) + 2 e Sn**(aq) + 2e 2 H*(aq) + 2 e → Hz(g) Pb2 (aq) + 2 e Sn2"(aq) + 2 e –→ Sn(s) NP*(aq) + 2 e –→ Ni(s) +0.77 Cu(s) +0.337 - 2 Hg(t) + 2 C(aq) → Sn2*(aq) +0.27 +0.15 0.00 → Pb(s) -0.126 -0.14 -0.25 Cd2"(aq) + 2 e – Cd(s) -0.40 Cr*(aq) + e → Cr*(aq) -0.408 Fe?"(aq) + 2 e – Fe(s) -0.44 Zn2"(aq) + 2 e -→Zn(s) -0.763 Cr*( aq) + 2 e – Cr(s) -0.91 AP (aq) + 3 e → Al(s) -1.66 Mg2 (aq) + 2 e –→ Mg(s) -2.37 V

Calculate the potential developed by a voltaic cell using the following reaction if all dissolved species are 0.011 M at 25 °C. 2 Fe2"(aq) + H2O2(aq) + 2 H¨(aq) → 2 Fe (aq) + 2 H,O(t) Acidic Solution Standard Electrode Potential, E(volts) H2O2{aq) + 2 H'(aq) + 2 e - 2 H2O(t) +1.77 Au(aq) + e →Au(s) +1.68 Au (aq) + 3 e-→ Au(s) Brz(8) + 2 e –2 Br(aq) NO: (aq) + 4 H(aq) + 3 e → NO(g) + 2 H20 Ag*(aq) + e +1.50 +1.08 +0.96 →Ag(s) +0.80 Hg2"(aq) + 2 e –→2 Hg(?) +0.789 Fe (aq) + e-→ Fe²"(aq) Cu²"(aq) + 2 e – Hg:Clz(s) + 2 e Sn**(aq) + 2e 2 H(aq) + 2 e → Hz(g) Pb2 (aq) + 2 e Sn2"(aq) + 2 e –→ Sn(s) NP(aq) + 2 e –→ Ni(s) +0.77 Cu(s) +0.337 - 2 Hg(t) + 2 C(aq) → Sn2(aq) +0.27 +0.15 0.00 → Pb(s) -0.126 -0.14 -0.25 Cd2"(aq) + 2 e – Cd(s) -0.40 Cr(aq) + e → Cr(aq) -0.408 Fe?"(aq) + 2 e – Fe(s) -0.44 Zn2"(aq) + 2 e -→Zn(s) -0.763 Cr( aq) + 2 e – Cr(s) -0.91 AP (aq) + 3 e → Al(s) -1.66 Mg2 (aq) + 2 e –→ Mg(s) -2.37 V

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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