Answer:
-372000 J or -372 KJ
Explanation:
We have the electrochemical reaction as;
Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)
We must first calculate the E∘cell from;
E∘cathode - E∘anode
E∘cathode = -0.44 V
E∘anode = -2.37 V
Hence;
E∘cell = -0.44 V -(-2.37 V)
E∘cell = 1.93 V
n= 2 since two electrons were transferred
F=96,500C/(mol e−)
ΔG∘=−nFE∘
ΔG∘= -( 2 * 96,500 * 1.93)
ΔG∘= -372000 J or -372 KJ