E=E ? ?2.303RT nF log 10 Q where E is the potential in volts, E ? is the standar
ID: 797766 • Letter: E
Question
E=E ? ?2.303RT nF log 10 Q where E is the potential in volts, E ? is the standard potential in volts, R=8.314J/(K?mol) is the gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred, F=96,500C/(mol e ? ) is the Faraday constant, and Q is the reaction quotient.
Consider the reactionMg(s)+Fe 2+ (aq)?Mg 2+ (aq)+Fe(s)at 47 ? C , where [Fe2+ ]= 2.80M and [Mg 2+ ]= 0.210M .
E=E ? ?2.303RT nF log 10 Q where E is the potential in volts, E ? is the standard potential in volts, R=8.314J/(K?mol) is the gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred, F=96,500C/(mol e ? ) is the Faraday constant, and Q is the reaction quotient. Consider the reactionMg(s)+Fe 2+ (aq)?Mg 2+ (aq)+Fe(s)at 47 ? C , where [Fe2+ ]= 2.80M and [Mg 2+ ]= 0.210M . What is the value for the reaction quotient, ,Q for the cell? Express your answer numerically. What is the value for the temperature, T , in kelvins? What is the value for n ?Express your answer as an integer and include the appropriate units (i.e. enter mol for moles). Calculate the standard cell potential forMg(s)+Fe 2+ (aq)?Mg 2+ (aq)+Fe(s) Express your answer to three significant figures and include the appropriate units.Explanation / Answer
(A) Q= [Mg2+(aq)]/[Fe2+(aq)]
=0.21/2.80
= 0.075
Part B What is the value for the temperature, T , in kelvins?
(B) T= 273+47 = 320 K
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).
(C) n=2 moles per one mole of reaction
(D)
Reduction: Fe2+ (aq) + 2 e– => Fe (s) E° = –0.45 V
Oxidation: Mg (s) => Mg2+ (aq) + 2 e– E° = +2.37 V
Overall: Mg(s) + Fe2+(aq) => Mg2+(aq) + Fe(s)
Eo(cell) = 2.37 - 0.45 = +1.92 V
RTnF= 8.314*320/(2*96500)= 0.0137849
E= Eo-2.303Rt/nF *logQ
= 1.92-(2.303*0.0137849)*log 0.075
= 1.92-(2.303*0.0137849)*(-1.125)
= 1.92-(-0.0357)
= 1.9557 V
So, E= 1.9557 Volts
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