Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M, [Al 3+] = 0.10 M, and the standard reduction potential is -1.66 V for Al 3+/Al and -0.45 V for Fe 2+/Fe.

1.18 V

Explanation:

The given cell is:

Half reactions for the given cell follows:

Oxidation half reaction:

Reduction half reaction:

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ?V

= standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.