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
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.