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Consider the following voltaic cell at 298K. Select only the statements that are TRUE. Cu (s) I Cu²+ (1.0 M) II A³+ (1.0 M) 1 Al (s) Al³+ (aq) + 3 e→→ Al(s) red = -1.66V Cu²+ (aq) + 2e → Cu(s) The anode is Cu(s). E E In the Nernst equation, the value of n = 5. Al³+ (aq) is reduced. red = 0.34V The reducing agent is Al³+ (aq).

Consider the following voltaic cell at 298K. Select only the statements that are TRUE. Cu (s) I Cu²+ (1.0 M) II A³+ (1.0 M) 1 Al (s) Al³+ (aq) + 3 e→→ Al(s) red = -1.66V Cu²+ (aq) + 2e → Cu(s) The anode is Cu(s). E E In the Nernst equation, the value of n = 5. Al³+ (aq) is reduced. red = 0.34V The reducing agent is Al³+ (aq).

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 64QAP: Consider a voltaic cell at 25°C in which the following reaction takes place:...
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Consider the following voltaic cell at 298K. Select only the statements that are TRUE. Cu (s) I Cu2+ (1.0 M) II A³+ (1.0 M) I AI (s) Al³+ (aq) + 3 e → Al(s) Ered = -1.66V Cu²+ (aq) + 2e → Cu(s) Ered = 0.34V In the Nernst equation, the value of n = 5. The anode is Cu(s). The reducing agent is Al³+ (aq). Al³+ (aq) is reduced.
Transcribed Image Text:Consider the following voltaic cell at 298K. Select only the statements that are TRUE. Cu (s) I Cu2+ (1.0 M) II A³+ (1.0 M) I AI (s) Al³+ (aq) + 3 e → Al(s) Ered = -1.66V Cu²+ (aq) + 2e → Cu(s) Ered = 0.34V In the Nernst equation, the value of n = 5. The anode is Cu(s). The reducing agent is Al³+ (aq). Al³+ (aq) is reduced.
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