Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 57 C , where [Fe2+]= 3.00

Question

Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 57 C , where [Fe2+]= 3.00 M and [Mg2+]= 0.110 M .

Part A What is the value for the reaction quotient, Q, for the cell?

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

              Q = 0.110 / 3.00

             Q = 0.037

part B)

T = 57 + 273 = 330 K

part C)

value of n = 2 . because here 2 electrons are transfered

part D)

Eocell = Eo red - Eo oxd

            = -0.44 - (-2.38)

           = 1.94 V

standard cell potential = 1.94 V

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