Question
Chemistry Question on Nernst Equation
Consider the cell Pt(s)∣H2(g,1atm)∣H+(aq,1M)∣∣Fe3+(aq),Fe2+(aq)Pt(s) When the potential of the cell is 0.712V at 298K, the ratio [Fe2+]/[Fe3+] is ______(Nearest integer) Given : Fe3++e−−Fe2+,EθFe3+,Fe2+Pt=0771 F2303RT=0.06V
Answer
The correct answer is 10.
Pt(s)∣H2(g,1atm)∣H+(aq,1M)∥Fe3+(aq),Fe2+(aq)∣Pt∣(s)
at anode H2⟶2H++2e−
At cathode Feaq3++e−⟶Feaq2+
E∘=EH2∣H+∘+EFe3+∣Fe2+∘=0⋅771V
E=E∘−10⋅06logFe3+Fe2+
0⋅712=(0+0⋅771)−10⋅06logFe3+Fe2+
logFe3+Fe2+=0⋅060⋅059≈1
Fe3+Fe2+=10