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Question: The standard electrode potentials (reduction) of Pt/Fe$^{3+}$, Fe$^{+2}$ and Pt/Sn$^{4+}$, Sn$^{+2}$...

The standard electrode potentials (reduction) of Pt/Fe3+^{3+}, Fe+2^{+2} and Pt/Sn4+^{4+}, Sn+2^{+2} are +0.77 V and 0.15 V respectively at 25°C. The standard EMF of the reaction Sn4+^{4+} + 2 Fe2+^{2+} →\rightarrow Sn2+^{2+} + 2Fe3+^{3+} is:-

A

-0.62 V

B

-0.92 V

C

+0.31 V

D

+0.85 V

Answer

-0.62 V

Explanation

Solution

To calculate the standard EMF (Electromotive Force) of the reaction, identify the reduction and oxidation half-reactions and their respective standard electrode potentials.

The given reaction is: Sn4+^{4+} + 2 Fe2+^{2+} →\rightarrow Sn2+^{2+} + 2Fe3+^{3+}

Breaking down the reaction into half-reactions:

  1. Reduction Half-reaction: Sn4+^{4+} is reduced to Sn2+^{2+}.
    Sn4+^{4+}(aq) + 2e−^{-} →\rightarrow Sn2+^{2+}(aq)
    The standard reduction potential given for this couple is E∘^{\circ}(Sn4+^{4+}/Sn2+^{2+}) = +0.15 V. This will be the potential at the cathode.

  2. Oxidation Half-reaction: Fe2+^{2+} is oxidized to Fe3+^{3+}.
    Fe2+^{2+}(aq) →\rightarrow Fe3+^{3+}(aq) + e−^{-}
    The standard electrode potential given for the Fe3+^{3+}/Fe2+^{2+} couple is a reduction potential: E∘^{\circ}(Fe3+^{3+}/Fe2+^{2+}) = +0.77 V.
    Since Fe2+^{2+} is being oxidized, we need the standard oxidation potential for Fe2+^{2+}/Fe3+^{3+}, which is the negative of the standard reduction potential:
    E∘^{\circ}(Fe2+^{2+}/Fe3+^{3+}) = -E∘^{\circ}(Fe3+^{3+}/Fe2+^{2+}) = -0.77 V. This will be the potential at the anode.

The standard EMF of the cell (Ecell∘^{\circ}_{cell}) can be calculated using the formula:
Ecell∘^{\circ}_{cell} = Ereduction∘^{\circ}_{reduction} (cathode) + Eoxidation∘^{\circ}_{oxidation} (anode)

Substituting the values:
Ecell∘^{\circ}_{cell} = E∘^{\circ}(Sn4+^{4+}/Sn2+^{2+}) + E∘^{\circ}(Fe2+^{2+}/Fe3+^{3+})
Ecell∘^{\circ}_{cell} = (+0.15 V) + (-0.77 V)
Ecell∘^{\circ}_{cell} = 0.15 V - 0.77 V
Ecell∘^{\circ}_{cell} = -0.62 V

Alternatively, using the formula where both potentials are reduction potentials:
Ecell∘^{\circ}_{cell} = Ecathode∘^{\circ}_{cathode} - Eanode∘^{\circ}_{anode}
In this reaction:

  • Sn4+^{4+}/Sn2+^{2+} is undergoing reduction, so it is the cathode. Ecathode∘^{\circ}_{cathode} = +0.15 V.
  • Fe3+^{3+}/Fe2+^{2+} is involved in oxidation (Fe2+^{2+} to Fe3+^{3+}), so it is the anode. Eanode∘^{\circ}_{anode} = +0.77 V (reduction potential of the anode couple).

Ecell∘^{\circ}_{cell} = E∘^{\circ}(Sn4+^{4+}/Sn2+^{2+}) - E∘^{\circ}(Fe3+^{3+}/Fe2+^{2+})
Ecell∘^{\circ}_{cell} = 0.15 V - 0.77 V
Ecell∘^{\circ}_{cell} = -0.62 V

Both methods yield the same result. The standard EMF of the reaction is -0.62 V.