Question

\(Pt(s) ∣ H2​(g)(1atm) ∣ H+(aq, [H+]=1)\, ∥\, Fe3+(aq), Fe2+(aq) ∣ Pt(s)\)
Given\( E^∘_{Fe^{3+}Fe^{2+}}\)\(=0.771V\) and \(E^∘_{H^{+1/2}H_2}​=0\,V,T=298K\)
If the potential of the cell is 0.712V, the ratio of concentration of \(Fe2+\) to \(Fe3+\) is

Answer 1

Correct Answer: 10

To solve for the ratio of concentrations of \(Fe^{2+}\) to \(Fe^{3+}\), we use the Nernst equation for the cell, which is given by:

\(E_{\text{cell}}=E^∘_{\text{cell}}-\frac{RT}{nF}\ln Q\)

For the cell reaction:

\(Fe^{3+}+e^- \rightleftharpoons Fe^{2+}\)

The standard cell potential is determined by:

\(E^∘_{\text{cell}}=E^∘_{Fe^{3+}/Fe^{2+}}-E^∘_{H^+/H_2}=0.771\,V-0\,V=0.771\,V\)

Given: \(E_{\text{cell}}=0.712\,V\), the number of electrons \(n=1\), and for reactions at \(298\,K\):

The Nernst equation simplifies to:

\(0.712=0.771-\frac{0.0591}{1}\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)\)

Rearranging for the concentration ratio:

\(\frac{0.0591}{1}\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=0.771-0.712=0.059\)

\(\log\left(\frac{[Fe^{2+}]}{[Fe^{3+}]}\right)=\frac{0.059}{0.0591}\approx1\)

Taking the antilogarithm:

\(\frac{[Fe^{2+}]}{[Fe^{3+}]}=10^1=10\)

Thus, the ratio of the concentration of \(Fe^{2+}\) to \(Fe^{3+}\) is 10, which fits the expected range of 10 to 10.