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# Answer to Question #56274 in General Chemistry for Megan

Question #56274
Calculate the cell potential for the following reaction as written at 25.00 degrees Celsius, given that [Cr^2+]= 0.839 M and [Fe^2+]= 0.0110 M.

Cr (s) + Fe^2+ (aq) <--> Cr^2+ (aq) + Fe (s)

Solution. To determine the cell potential, it is necessary to calculate electrode potentials. For this purpose we find values of standard electrode potentials of Cr2 +/Cr system (-0,744 B) and Fe2 +/Fe (-0,440 B) in the reference book, and then we calculate values of potentials on Nernst's equation:
E(Сr2+/Cr) = -0,744 + (0,059/2)*lg(0,839) = -0,744 + (0,059/2)(-0.076) = -0,746 V,
E(Fe2+/Fe) = -0,13 + (0,059/2)*lg(0,0110) = -0,13 + (0,059/2)(-1.959)= -0,188 V.
Element EMF: E = E (Fe2 +/Fe) - E (Zn2 +/Zn) =-0,188 - (-0,746) = 0,558 V.

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