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