Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt. The E^ _ Fe^ 3+ /Fe^ 2+ in Volt at 298 K is given by:

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 3Y-2X

The half-reactions and their standard Gibbs free energy changes are given by:

Fe^{2+} + 2e^{-} \rightarrow Fe

\Delta G^{\ominus}_{1} = -2FX

Fe^{3+} + 3e^{-} \rightarrow Fe

\Delta G^{\ominus}_{2} = -3FY

The required half-reaction is:

Fe^{3+} + e^{-} \rightarrow Fe^{2+}

\Delta G^{\ominus}_{3} = -1FE^{\ominus}_{Fe^{3+}/Fe^{2+}}

This reaction is obtained by subtracting the first reaction from the second reaction:

\Delta G^{\ominus}_{3} = \Delta G^{\ominus}_{2} - \Delta G^{\ominus}_{1}

-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY - (-2FX)

-FE^{\ominus}_{Fe^{3+}/Fe^{2+}} = -3FY + 2FX

E^{\ominus}_{Fe^{3+}/Fe^{2+}} = 3Y - 2X

Answer:

3Y-2X

Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt....