Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 69 C , where [Fe2+]= 3.20

ID: 953091 • Letter: C

Question

Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 69 C , where [Fe2+]= 3.20 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.

THANK YOU!!!

part A ) Q = [Mg+2 ] /[Fe+2 ] = 0.110/3.2 = 0.0343

part B ) T = 273.15 + 69 = 342.15 K

PART C) n = no of electrons = 2

part D ) Ecell = Eocell - 2.303/2RTlogQ

= 2.37 + (-0.44) - 0.04log(0.0343 )

= 1.93 + 0.0585 = 1.988 Volt

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