[Thuvaraja1]

one-half cell of an electrolytic cell has a lanthanum electrode in a 1 mole/L lanthanum  hyposulfite solution. The other half-cell has a magnesium electrode in a 1 mole/L magnesium hyposulfite solution. How would adding more lanthanum ions to the cathode half-cell affect the cell voltage?

La⁺³ + 3e^- → La       E(cathode) = -2.379V

Mg → Mg⁺² +2e^-      E(anode) = -2.372V

E(cell) = E(cathode) - E(anode)
         = -2.379V - (-2.372V)
         = -0.007V

I thought about it and I think that by adding more lanthanum ions, your increasing it's concentration, so its reduction potential would increase. That means that the cell voltage should increase, but I'm not really sure.

[IITian]

The E^o(Cathode) is constant, but E (Cathode) depends on the concentration. Thus, the net effect can be obtained by Nernst Equation.