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)
please help, i have gotten it wrong every time I tried, this is the fourth time
Cr (s) + Fe^2+ (aq) <--> Cr^2+ (aq) + Fe (s)
please help, i have gotten it wrong every time I tried, this is the fourth time
Expert's answer
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.
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.
