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A concentration cell based on the following half reaction at 316 K Mg2+ + 2e- –→ Mg; SRP = -2.37 is made from two 1.00 L solutions of Mg2+, 1.34 M and 0.272 M; two 100.0 g pieces of Mg metal, a wire, and a salt bridge. Under these conditions the potential of the cell is 0.02171 V. How much Mg metal will need to be oxidized for the potential of the cell to decrease by 0.011725 V? Give your answer to 3 decimal places. (g) 1.089 X 0%

A concentration cell based on the following half reaction at 316 K Mg2+ + 2e- –→ Mg; SRP = -2.37 is made from two 1.00 L solutions of Mg2+, 1.34 M and 0.272 M; two 100.0 g pieces of Mg metal, a wire, and a salt bridge. Under these conditions the potential of the cell is 0.02171 V. How much Mg metal will need to be oxidized for the potential of the cell to decrease by 0.011725 V? Give your answer to 3 decimal places. (g) 1.089 X 0%

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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A concentration cell based on the following half reaction at 316 K Mg²+ + 2e- → Mg; SRP = -2.37 is made from two 1.00 L solutions of Mg?+, 1.34 M and 0.272 M; two 100.0 g pieces of Mg metal, a wire, and a salt bridge. Under these conditions the potential of the cell is 0.02171 V. How much Mg metal will need to be oxidized for the potential of the cell to decrease by 0.011725 V? Give your answer to 3 decimal places. (6) 1.089 x 0% This question is complete and cannot be answered again. Correct answer (g) = 6.10181
Transcribed Image Text:A concentration cell based on the following half reaction at 316 K Mg²+ + 2e- → Mg; SRP = -2.37 is made from two 1.00 L solutions of Mg?+, 1.34 M and 0.272 M; two 100.0 g pieces of Mg metal, a wire, and a salt bridge. Under these conditions the potential of the cell is 0.02171 V. How much Mg metal will need to be oxidized for the potential of the cell to decrease by 0.011725 V? Give your answer to 3 decimal places. (6) 1.089 x 0% This question is complete and cannot be answered again. Correct answer (g) = 6.10181
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