chemistry
posted by nj on .
A voltaic cell at 25oC consists of Mn/Mn2+ and Cd/Cd2+ halfcells with the initial concentrations [Mn2+] = 0.100 M and [Cd2+] = 0.0100 M. Use the Nernst equation to calculate E for this cell.
Cd+2(aq) + 2e = Cd(s) . . . . . . Eo = 0.40 V
Mn+2(aq) + 2e = Mn(s) . . . . . . . Eo = 1.18 V

The overall reaction is:
Mn(s) + Cd^2+(aq) > Mn^2+(aq) + Cd(s)
(Mn is higher on the activity series than Cd)
Eo(cell) = Eo(Cd) Eo(Mn) with 1M solutions at 298K
Eo(cell) = 40v  (1.18v) = 0.78v
With the given concentrations,
E(cell) = Eo(cell)  (0.059/n)(logQ),
where Q = [Mn^2] / [Cd^2+], and
n = 2
Substitute and solve for E(cell) in the Nernst Equation above.