Question

Calculate the voltage generated in the following cell Mn|Mn2+||Fe2+|Fe when iron rod is immersed in 1.0 M FeSO4 solution and Mn rod is immersed in 0.1 M MnSO4 solution. Given E0Fe2+/Fe = -0.40 V and E0Mn2+/Mn = -1.18 V. Write the cell representation and reactions.

Answer 1

$Mn=Mn^{2+}+2e^- (Anode) E_A=-1.18V$

$Fe^{2+}+2e^-=Fe(Cathode) E_C=-0.40V$

$E=E_A-E_C+\frac{0.059}{2}lg\frac{a_{Fe^{2+}}}{a_{Mn^{2+}}}=-0.40+1.18+\frac{0.059}{2}lg\frac{1.0}{0.1}=0.81V$

Answer E=0.81V