Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 71 C , where [Fe2+]= 2.90

Question

Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 71 C , where [Fe2+]= 2.90 M and [Mg2+]= 0.110 M . Part A What is the value for the reaction quotient, Q, for the cell? Express your answer numerically. Part B What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units. Part C What is the value for n? Express your answer as an integer and include the appropriate units (i.e. enter mol for moles). Part D Calculate the standard cell potential for Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Part A) Q = [Mg2+]/[Fe2+] = 0.110/2.90 = 0.0379

Part B) T = 71 + 273.15 = 344.15 K

Part C) n = 2 mol, since 2 electrons are transferred in the redox reaction

Part D) At equilibrium E0 = RT/nF ln Q = 0.0591/2 (ln 0.0379) = 0.0295 ln 0.0379 = - 0.096 V

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