Question

Consider the following standard electrode potential values:

\[\ce{Sn^{2+}_{ (aq)} + 2e^- -> Sn_{(s)}}\]; E⁰ = −0.14 V

\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\]; E⁰ = +0.77 V

What is the cell reaction and potential for the spontaneous reaction that occurs?

Options

• \[\ce{2Fe^{2+}_{ (aq)} + Sn^{2+}_{ (aq)} -> 2Fe^{3+}_{ (aq)} + Sn_{(s)}}\]; E⁰ = −0.91 V

• \[\ce{2Fe^{3+}_{ (aq)} + Sn_{(s)} -> 2Fe^{2+}_{ (aq)} + Sn^{2+}_{ (aq)}}\]; E⁰ = +0.91 V

• \[\ce{2Fe^{2+}_{ (aq)} + Sn^{2+}_{ (aq)} -> 2Fe^{3+}_{ (aq)} + Sn_{(s)}}\]; E⁰ = +0.91 V

• \[\ce{2Fe^{3+}_{ (aq)} + Sn_{(s)} -> 2Fe^{3+}_{ (aq)} + Sn^{2+}_{ (aq)}}\]; E⁰ = +1.68 V

Answer 1

\[\ce{2Fe^{3+}_{ (aq)} + Sn_{(s)} -> 2Fe^{2+}_{ (aq)} + Sn^{2+}_{ (aq)}}\]; E⁰ = +0.91 V

Explanation:

Cell potential

\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]

= +0.77 − (−0.14)

= +0.91 V

The cell reaction will be

\[\ce{2Fe^{3+}_{ (aq)} + Sn_{(s)} -> 2Fe^{2+}_{ (aq)} + Sn^{2+}_{ (aq)}}\]