Compare the calculations from Step 2 and Step 3. What effect does changing zero on the scale of reduction potentials have on: a. Reduction potetials? (Individual) b. Cell potentials? (Voltaic Cell System)
Compare the calculations from Step 2 and Step 3. What effect does changing zero on the scale of reduction potentials have on: a. Reduction potetials? (Individual) b. Cell potentials? (Voltaic Cell System)
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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STEP 2 AND STEP 3 HAVE ALREADY ANSWERS. PLEASE ANSWER THE QUESTION BELOW:
Compare the calculations from Step 2 and Step 3. What effect does changing zero on the scale of reduction potentials have on:
a. Reduction potetials? (Individual)
b. Cell potentials? (Voltaic Cell System)
Transcribed Image Text:Step 2
Use the given standard reduction potentials to calculate the standard cell potentials Explanation
Enet-Ec-Ea using the SPONTANEOUS
for the following redox reactions:
rules?
(a) 2Li(s) + 2H(aq) → 2Li (aq) + H₂(g)
(b) 2Al(s) + 3F₂(g) → 2A1³+ (aq) + 6F (aq)
(c) 2FeCl3(aq) + H₂(g) → 2FeCl₂(aq) + 2HCl(aq)
(d) Al(NO3)3(aq) + 3Li(s)→ 3LiNO3(aq) + Al(s)
Solutions
(a) 2Li(a) +2H* (aq) →2Li (aq) + H₂(g)
Enet = +3.04 V
Explanation
Oxidation reaction: 2Li(a) →2Li* (eg) +2e
Reduction reaction: 2H(aq) +2e → H₂(g)
Enet = 0.0 V-(-3.04V)
= +3.04 V
E = -3.04 V
E=0.00 V
(b) 2Al(s) + 3F2(g) → 2Al³+ (
Enet = +4.53 V
+6F (sq)
(aq)
Oxidation reaction: 2Al(a) 2Al³+ (aq) + Ge
Reduction reaction: 3F2(g) +6e →6F (aq)
Enet 2.87 V-(-1.66V)
= +4.53 V
(c) 2FeCl3(aq) + H2(g) →2FeCl2(aq) + 2HCl(aq)
Enet = +0.77 V
Explanation
Oxidation reaction: H₂(g) → 2H*(aq) +2e
Reduction reaction:2Fe³+, (eg) + 2€* →2Fe²+ (aq)
Enet = 0.77 V-0.0 V
= +0.77 V
(d) Al(NO3)3(aq) +3Li(a) →3LINO3(aq) + Al(a)
Enet = +1.38 V
Explanation
Oxidation reaction: 3Li(a)
Reduction reaction: Al³+ (aq) + 3€¯ → Al(a)
3e
Enet-1.66 V-(-3.04V)
= -1.66 V+3.04 V
= +1.38 V
E = -1.66V
E = +2.87V
E=0.0V
E = +0.77 V
3Li (aq) +3e E= -3.04 V
E = -1.66 V
Transcribed Image Text:Step 3
b)
Repeat your calculations using the new, adjusted reduction potentials. (with New
A³+ /Al being the reference cell)
(a) 2Li(s) + 2H* (aq) → 2Li (aq) + H₂(g)
(b) 2Al(s) + 3F₂(g) → 2A1³+ (aq) + 6F (aq)
(c) 2FeCl3(aq) + H₂(g) → 2FeCl₂(aq) + 2HCl(aq)
(d) Al(NO3)3(aq) + 3Li(s) → 3LiNO3(aq) + Al(s)
Solutions
(a) Anode: 2L
2₁² +2e-
H₂
Cathode: 2H + 2 e-
Ecell Ecathode Eanode
Ent/H₂-ELit/Li
Ecell = 1.66 - (-1.38)
Ecell =
3.04 V
=
EL/L= -1.38 V
EH/H₂ = 1.66 V
Anode: 2A1+
Cathode: 3F₂ +6e² → 6F²
→ 2A1³+ + 6e-
ECathode - Eanode
E Cell
Ecell
(d) Anode:
Ecell
(C) Cathode 2 FeCl3+2€->2FeCl₂
Anode: H₂
= 4.53-0
= 4.53 V
Ecell
Ecell
2
Ecell
Z
Ecathode Eanode
Effet-EH/H₂
→ 2HCl +2e²° E₁H+/H₂
= 2.43 -1.66
0.77 V
3L3L NO3 + 3e-
+ 3e
AI
Cathode: Al(NO3)3
Ecell
Ecathode -Eanode
EA1³4/A1 = 0
EF₂/F-
z 0 - (-1.38)
= 1.38 V
= 4.53 V
EFe³+/Fe²+ = 2.43 V
= 1.66 V
= -1.38 V
ELi Li
EA13+/A1 = 0
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