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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: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.101QE: At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction...
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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)

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 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
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
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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