= 2.26A voltaio cell employs the following redox reaction:2 Fe (aq) +3 Mg (s) 2 Fe (s) +3 Mg (aq)Calculate the cell potential at 25 °C under each of thefollowing conditions.In this solution, Ecel < E°cell because there are more(Q > 1). Therefore, the reaction will proceed to the left.Part C[Fe"] = 2.85 M ; [Mg?+] = 2.1x10-3 MExpress your answer in units of volts.Ecell=V

Given, 2Fe3+(aq) + 3Mg(s) ➝ 2Fe(s) + 3Mg2+(aq) Temperature (T) = 25°C [Fe3+] = 2.85 M; [Mg2+] = 2.1…

= 2.26 A voltaio cell employs the following redox reaction: 2 Fe (aq)+3 Mg (s)2 Fe (s) +3 Mg²+ (aq) Calculate the cell potential at 25 °C under each of the following conditions. In this solution, Ecell < E°cell because there are more (Q > 1). Therefore, the reaction will proceed to the left. Part C [Fe] = 2.85 M ; [Mg²+] = 2.1x10-3 M Express your answer in units of volts. Ecell=

= 2.26 A voltaio cell employs the following redox reaction: 2 Fe (aq)+3 Mg (s)2 Fe (s) +3 Mg²+ (aq) Calculate the cell potential at 25 °C under each of the following conditions. In this solution, Ecell < E°cell because there are more (Q > 1). Therefore, the reaction will proceed to the left. Part C [Fe] = 2.85 M ; [Mg²+] = 2.1x10-3 M Express your answer in units of volts. Ecell=

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter17: Electrochemistry And Its Applications

Section: Chapter Questions

Problem 22QRT

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At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction PbC2O4(s)+2ePb(s)+C2O42(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/PbC2O4 electrode in a 0.025 M solution of Na2C2O4.
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