The standard electrode potentials (reduction) of Pt/Fe3+, Fe+2 and Pt/Sn4+, Sn+2 are +0.77 V and 0.15 V
respectively at 25°C. The standard EMF of the reaction
\(Sn^{4+} + 2Fe^{2+} \rightarrow Sn^{2+} + 2Fe^{3+}\) is
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 0.62 V
To determine the standard EMF of the reaction $\text{Sn}^{4+} + 2\text{Fe}^{2+} \rightarrow \text{Sn}^{2+} + 2\text{Fe}^{3+}$, we can use the standard electrode potentials and the Nernst equation.
Given:
Standard electrode potential (reduction) of Pt/Fe$^{3+}$, Fe$^{2+}$ ($E^\circ(\text{Pt/Fe}^{3+}, \text{Fe}^{2+}$)) = +0.77 V
Standard electrode potential (reduction) of Pt/Sn$^{4+}$, Sn$^{2+}$ ($E^\circ(\text{Pt/Sn}^{4+}, \text{Sn}^{2+}$)) = 0.15 V
The standard EMF ($E^\circ$) of the reaction can be calculated by subtracting the standard electrode potential of the reactants from the standard electrode potential of the products: $E^\circ = E^\circ(\text{Pt/Sn}^{4+}, \text{Sn}^{2+}) - E^\circ(\text{Pt/Fe}^{3+}, \text{Fe}^{2+})$
$E^\circ = 0.15 \, \text{V} - (+0.77 \, \text{V})$
$E^\circ = -0.62 \, \text{V}$
Therefore, the standard EMF of the reaction $\text{Sn}^{4+} + 2\text{Fe}^{2+} \rightarrow \text{Sn}^{2+} + 2\text{Fe}^{3+}$ is (1) $-0.62 \, \text{V}$.