2. Calculate the standard free energy change for the following reaction
Ni2+ (aq) + Fe(s) → Ni(s) +Fe2+ (aq)
If E°(Ni2+/Ni) = -0.25 V and E°( Fe2+/Fe) = -
0.44 V (Faraday constant 96,500 J/.mol) *
(2.5 Points)

-36.67 KJ

Explanation:

Now, we ,must find the  E°cell as follows;

E°cell= E°cathode -  E°anode

E°cathode= -0.25 V

E°anode = -  0.44 V

E°cell= -0.25 -(-  0.44) = 0.19 V

Equation of the reaction; Ni2+ (aq) + Fe(s) → Ni(s) +Fe2+ (aq)

Hence, n=2, there were two electrons transferred.

ΔG=-nFE°cell

ΔG= -(2 * 96,500 * 0.19)

ΔG= -36670 J

ΔG= -36.67 KJ

particle   p is an electron and particle q is a neutron

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >