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Science Chemistry Chemistry: Principles and Practice

Calculate the potential for each of the voltaic cells in Exercise 18.43 when the concentrations of the soluble species and gas pressures are as follows: (a) [Cu 2+ ] = 0.050 M , [Ni 2+ ] = 1.40 M (a) P C l 2 = 32 0 torr (b) [ I − ] = 0.00 1 0 M , P Cl 2 = 0. 3 00 atm , [ Cl - ] = 0. 6 0 M (a) Cu 2+ (aq) + Ni(s) → Cu(s) + Ni 2+ (aq) (b) 2Ag(s) + Cl 2 (g) → 2AgCl(s) (c) Cl 2 (g) + 2I − (aq) → 2Cl − (aq) = I 2 (s)

Calculate the potential for each of the voltaic cells in Exercise 18.43 when the concentrations of the soluble species and gas pressures are as follows: (a) [Cu 2+ ] = 0.050 M , [Ni 2+ ] = 1.40 M (a) P C l 2 = 32 0 torr (b) [ I − ] = 0.00 1 0 M , P Cl 2 = 0. 3 00 atm , [ Cl - ] = 0. 6 0 M (a) Cu 2+ (aq) + Ni(s) → Cu(s) + Ni 2+ (aq) (b) 2Ag(s) + Cl 2 (g) → 2AgCl(s) (c) Cl 2 (g) + 2I − (aq) → 2Cl − (aq) = I 2 (s)

Solution Summary: The author explains that the value of the reaction must be determined with the given concentrations.

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: Cengage Learning

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Textbook Question

Chapter 18, Problem 18.61QE

Calculate the potential for each of the voltaic cells in Exercise 18.43 when the concentrations of the soluble species and gas pressures are as follows:

(a) [Cu²⁺] = 0.050 M, [Ni²⁺] = 1.40 M
(a) $P C l 2 = 32 0 torr$
(b) $[ I − ] = 0.00 1 0 M , P Cl 2 = 0. 3 00 atm , [ Cl - ] = 0. 6 0 M$
(a) Cu²⁺ (aq) + Ni(s) → Cu(s) + Ni²⁺ (aq)
(b) 2Ag(s) + Cl₂(g) → 2AgCl(s)
(c) Cl₂(g) + 2I⁻(aq) → 2Cl⁻(aq) = I₂(s)

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Chapter 18, Problem 18.60QE

Chapter 18, Problem 18.62QE

Chapter 18 Solutions

Chemistry: Principles and Practice

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