Question

What is the cell potential (emf, in V) of this cell at 25oC? Mn | Mn2+(0.0264 M) ǁ H2SO4 (0.284 M) |SO2 (1 atm) |Pt(s)

Answer 1

SO42− + 4 H+ + 2 e− ----> SO2(aq) + 2 H2O E0 = +0.17 V

Mn2+ + 2 e− ---> Mn(s)    E0 = −1.185 v

cell reaction: Mn(s) + SO42−(aq) + 4 H+(aq) ----> Mn2+(aq) + SO2(aq) + 2 H2O(l)

E0cell = Ecathode - Eanode

       = (0.17)-(-1.185)

       = 1.355 v

Ecell = E0cell - (0.0591/n)log([Mn2+]*pSO2/[SO42−][H+]^4)

      = 1.355-(0.0591/2)log((0.0264*1/(0.284*(0.284*2)^4)))

      = 1.36 V