Solveeit Logo

Question

Chemistry Question on Nernst Equation

Consider the cell Pt(s)H2(g,1atm)H+(aq,1M)Fe3+(aq),Fe2+(aq)Pt(s)Pt _{( s )}\left| H _2( g , 1 atm )\right| H ^{+}( aq , 1 M )|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) Pt ( s ) When the potential of the cell is 0.712V0.712 \, V at 298K298 \,K, the ratio [Fe2+]/[Fe3+]\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right] is ______(Nearest integer) Given : Fe3++eFe2+,EθFe3+,Fe2+Pt=0771Fe ^{3+}+ e ^{-}- Fe ^{2+}, E ^\theta Fe ^{3+}, Fe ^{2+} Pt =0771 2303RTF=0.06V\frac{2303 RT }{ F }=0.06 \,V

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⋅06​logFe3+Fe2+​
0⋅712=(0+0⋅771)−10⋅06​logFe3+Fe2+​
logFe3+Fe2+​=0⋅060⋅059​≈1
Fe3+Fe2+​=10