What is the cell potential for the reaction Mg(s) + Fe^2+ yields Mg^2+ + Fe(s)?
First fine Ezero for the cell. Mg yields 2e plus Mg++ +2.37 volts Fe++ +2e yields Fe -.44 volts Add these voltages to get E zero for a standard cell. Comes out +l.93 volts Now you use the Nernst equation to find the voltage of a cell with non standard concentrations. and that is. E = Eo – RT over nF times lnQ E is what you; are looking for, cell voltage R= 8.31 joules/mole/degree K n = number of electrons being transfered which is 2 T= Kelvin temp. 273 plus 77 F= 9.648 x l0^4 coulombs per Faraday Eo is what you already found, l.93 volts Q = the quotient of your concentration of ions. product over reactant, (Mg++) over (Fe++) each raised to the power of their coefficient which in this case is l. so Q = (.301 Mg++) over (3.50 Fe++) Now just enter all these numbers into the Nernst equation and solve for E. I would do it but I am worn out from doing Yahoo answers all morning.
