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Question: Consider the following standard electrode potential values $Sn^{2+}_{(aq)} + 2e^- \rightarrow Sn_{(...

Consider the following standard electrode potential values

Sn(aq)2++2e−→Sn(s)Sn^{2+}_{(aq)} + 2e^- \rightarrow Sn_{(s)} E°= -0.14 V Fe(aq)3++e−→Fe(aq)2+Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)} E° = +0.77 V

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

A

2Fe(aq)2++Sn(aq)2+→2Fe(aq)3++Sn(s)2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)} \rightarrow 2Fe^{3+}_{(aq)} + Sn_{(s)}; E° = -0.91 V

B

2Fe(aq)3++Sn(s)→2Fe(aq)2++Sn(aq)2+2Fe^{3+}_{(aq)} + Sn_{(s)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}; E° = +0.91 V

C

2Fe(aq)2++Sn(aq)2+→2Fe(aq)3++Sn(s)2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)} \rightarrow 2Fe^{3+}_{(aq)} + Sn_{(s)}; E° = +0.91 V

D

2Fe(aq)3++Sn(s)→2Fe(aq)2++Sn(aq)2+2Fe^{3+}_{(aq)} + Sn_{(s)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}; E° = +1.68 V

Answer

2Fe(aq)3++Sn(s)→2Fe(aq)2++Sn(aq)2+2Fe^{3+}_{(aq)} + Sn_{(s)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}; E° = +0.91 V

Explanation

Solution

To determine the spontaneous cell reaction and its potential, we first compare the standard reduction potentials of the given half-reactions:

  1. Sn(aq)2++2e−→Sn(s)Sn^{2+}_{(aq)} + 2e^- \rightarrow Sn_{(s)}; E∘=−0.14 VE^\circ = -0.14 \, V
  2. Fe(aq)3++e−→Fe(aq)2+Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}; E∘=+0.77 VE^\circ = +0.77 \, V

For a spontaneous reaction, the species with the higher standard reduction potential will be reduced (act as cathode), and the species with the lower standard reduction potential will be oxidized (act as anode).

Comparing the values: +0.77 V>−0.14 V+0.77 \, V > -0.14 \, V.
Therefore, Fe3+Fe^{3+} will be reduced to Fe2+Fe^{2+}, and SnSn will be oxidized to Sn2+Sn^{2+}.

Reduction half-reaction (at cathode):
Fe(aq)3++e−→Fe(aq)2+Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}; Ecathode∘=+0.77 VE^\circ_{cathode} = +0.77 \, V

Oxidation half-reaction (at anode):
The reverse of the Sn reduction reaction:
Sn(s)→Sn(aq)2++2e−Sn_{(s)} \rightarrow Sn^{2+}_{(aq)} + 2e^-; Eanode∘=−(−0.14 V)=+0.14 VE^\circ_{anode} = -(-0.14 \, V) = +0.14 \, V

To obtain the overall cell reaction, we need to balance the electrons. Multiply the reduction half-reaction by 2:
2Fe(aq)3++2e−→2Fe(aq)2+2Fe^{3+}_{(aq)} + 2e^- \rightarrow 2Fe^{2+}_{(aq)}

Now, add the balanced half-reactions:
Sn(s)+2Fe(aq)3++2e−→Sn(aq)2++2e−+2Fe(aq)2+Sn_{(s)} + 2Fe^{3+}_{(aq)} + 2e^- \rightarrow Sn^{2+}_{(aq)} + 2e^- + 2Fe^{2+}_{(aq)}
Cancelling the electrons, the overall spontaneous cell reaction is:
2Fe(aq)3++Sn(s)→2Fe(aq)2++Sn(aq)2+2Fe^{3+}_{(aq)} + Sn_{(s)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}

Now, calculate the standard cell potential (Ecell∘E^\circ_{cell}):
Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} (using standard reduction potentials)
Ecell∘=E∘(Fe3+/Fe2+)−E∘(Sn2+/Sn)E^\circ_{cell} = E^\circ(Fe^{3+}/Fe^{2+}) - E^\circ(Sn^{2+}/Sn)
Ecell∘=(+0.77 V)−(−0.14 V)E^\circ_{cell} = (+0.77 \, V) - (-0.14 \, V)
Ecell∘=+0.77 V+0.14 VE^\circ_{cell} = +0.77 \, V + 0.14 \, V
Ecell∘=+0.91 VE^\circ_{cell} = +0.91 \, V

Alternatively, Ecell∘=Ereduction∘+Eoxidation∘E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation}
Ecell∘=(+0.77 V)+(+0.14 V)E^\circ_{cell} = (+0.77 \, V) + (+0.14 \, V)
Ecell∘=+0.91 VE^\circ_{cell} = +0.91 \, V

Comparing this with the given options, option (b) matches both the cell reaction and the potential.