posted by nj .
A voltaic cell at 25oC consists of Mn/Mn2+ and Cd/Cd2+ half-cells 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.