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!!!

Answer 1

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 = E^ocell - 2.303/2RTlogQ

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

                     = 1.93 + 0.0585 = 1.988 Volt