Question

A voltaic cell is made by connecting two half cells represented by half equations below:

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

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

Which statement is correct about this voltaic cell?

Options

• Fe²⁺ is oxidised and the voltage of the cell is −0.91 V.

• Sn is oxidised and the voltage of the cell is 0.91 V.

• Fe²⁺ is oxidised and the voltage of the cell is 0.91 V.

• Sn is oxidised and the voltage of the cell is 0.63 V.

Answer 1

Sn is oxidised and the voltage of the cell is 0.91 V.

Explanation:

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

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

The reduction potential of \[\ce{Fe^{3+}_{ (aq)}}\] is higher than that of \[\ce{Sn^{2+}_{ (aq)}}\].

As a result, sn_(s) will be oxidised and Fe³⁺ will be decreased.

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

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

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

E = 0.77 − (−0.14)

E = 0.91 V