Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 89 C , where [Fe2+]= 2.90

Question

Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 89 C , where [Fe2+]= 2.90 M and [Mg2+]= 0.210 M . Part A What is the value for the reaction quotient, Q, for the cell? Express your answer numerically. Q =

Part B What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units. T =

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). n =

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. E =

Explanation / Answer

Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)

Q = [Mg+2]/[Fe+2]

     = 0.21/2.9 = 0.0724

part-B

T = 89C0 = 89+273 = 362K

part-C

Mg-(s)---------------> Mg+2 (aq) + 2e-

Fe+2(aq) + 2e- -------> Fe(s0

--------------------------------------------------

Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)

n   = 2

part-C

Mg-(s)---------------> Mg+2 (aq) + 2e-   E0 = 2.38V

Fe+2(aq) + 2e- -------> Fe(s)              E0 = -0.41

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Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)   E0Cell = 1.97V

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