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 SolutionStandard Electrode Potential, E (volts)H₂O₂(aq) + 2 H*(aq) + 2 e →→ 2 H₂O(l)+1.77Aut(aq) + e — Au(s)+1.68+1.50Au"(aq) + 3e — Au(s)Brz(t) + 2 e2 Br(aq)+1.08NO3(aq) + 4 H(aq) + 3 e NO(g) + 2 H₂O+0.96Ag*(aq) + e → → Ag(s)+0.80Hgz²(aq) + 2 e2 Hg(l)+0.789Fe³(aq) + e-→→→→→ Fe²(aq)+0.77Cu²(aq) + 2 e - Cu(s)+0.337HgzClz(s) + 2 e2 Hg(t) + 2 Cl(aq)+0.27Sn(aq) + 2e →→→ Sn²+ (aq)+0.150.002 H*(aq) + 2 e →→→H₂(g)Pb²(aq) + 2e →→→→ Pb(s)-0.126Sn²(aq) + 2 e-> Sn(s)-0.14Ni²+ (aq) + 2 e → Ni(s)-0.25-0.40Cd²+ (aq) + 2 e → Cd(s)Cr³(aq) + eCr²+ (aq)-0.408Fe²(aq) + 2 e→→→→> Fe(s)-0.44Zn² (aq) + 2 e→→→→→→ Zn(s)-0.763Cr²(aq) + 2 e Cr(s)-0.91Al³+ (aq) + 3 e-> Al(s)-1.66Mg²+ (aq) + 2 e →→→Mg(s)-2.37V

Given cell reaction is 2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) ---> 2 Fe3+(aq) + 2 H2O(l) Temperature…

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) -

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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Question

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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