Calculate the potential developed by a voltaic cell using the following reaction if alldissolved species are 0.016 M at 25 °C.2 Fe²+ (aq) + H₂O₂(aq) + 2 H†(aq) → 2 Fe³+ (aq) + 2 H₂O(l)Acidic SolutionStandard Electrode Potential, E (volts)+1.77H₂O₂(aq) + 2 H*(aq) + 2 e → 2 H₂O(l)Au*(aq) + e —>Au(s)+1.68+1.50Au3'(aq) +3e — Au(s)Br₂(l) +2 e2 Br(aq)+1.08NO3(aq) + 4 H*(aq) + 3 e+0.96Ag*(aq) + e →→→Ag(s)+0.80Hg₂² (aq) + 2 e→→→→→2 Hg(l)+0.789Fe³(aq) + e-Fe²(aq)+0.77Cu²+ (aq) + 2 e→→→→→ Cu(s)+0.337HgzClz(s) + 2 e+0.27Sn(aq) + 2e-→+0.152 H*(aq) + 2 e→→→→→ H₂(g)0.00Pb²+ (aq) + 2 e→→→→→→ Pb(s)-0.126Sn²+ (aq) + 2 e→→→→→Sn(s)-0.14Ni2+ (aq) + 2 e-Ni(s)-0.25Cd²+ (aq) + 2 eCd(s)-0.40Cr³+ (aq) + e-Cr²+ (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)-2.37NO(g) + 2 H₂O2 Hg(t) + 2 Cl(aq)Sn²+ (aq)PreviousNext

Form given table E°fe3+/fe2+ = 0.77 V E°H2O2 = 1.77 Now, E°cell = E°c - E°a = 1.77 - 0.77 = 1.00…

→2 Hg(t) Fe²(aq) → Cu(s) →2 Hg(t) + 2 Cl(aq) - Sn²+ (aq) → H₂(g) → Pb(s) → Sn(s) - Ni(s) → Cd(s) Cr²+ (aq) Fe(s) +0.789 +0.77 +0.337 +0.27 +0.15 0.00 -0.126 -0.14 -0.25 -0.40 -0.408 -0.44

→2 Hg(t) Fe²(aq) → Cu(s) →2 Hg(t) + 2 Cl(aq) - Sn²+ (aq) → H₂(g) → Pb(s) → Sn(s) - Ni(s) → Cd(s) Cr²+ (aq) Fe(s) +0.789 +0.77 +0.337 +0.27 +0.15 0.00 -0.126 -0.14 -0.25 -0.40 -0.408 -0.44

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 0.016 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)
+1.77
H₂O₂(aq) + 2 H*(aq) + 2 e → 2 H₂O(l)
Au*(aq) + e —>Au(s)
+1.68
+1.50
Au3'(aq) +3e — Au(s)
Br₂(l) +2 e
2 Br(aq)
+1.08
NO3(aq) + 4 H*(aq) + 3 e
+0.96
Ag*(aq) + e →→→Ag(s)
+0.80
Hg₂² (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 e
+0.27
Sn(aq) + 2e-
→
+0.15
2 H*(aq) + 2 e→→→→→ H₂(g)
0.00
Pb²+ (aq) + 2 e→→→→→→ Pb(s)
-0.126
Sn²+ (aq) + 2 e→→→→→Sn(s)
-0.14
Ni2+ (aq) + 2 e-
Ni(s)
-0.25
Cd²+ (aq) + 2 e
Cd(s)
-0.40
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
NO(g) + 2 H₂O
2 Hg(t) + 2 Cl(aq)
Sn²+ (aq)
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