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  • In a cell, the following reactions take place <img alt="\begin{aligned} &\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \quad \mathrm{E}_{\mathrm{Fe}^{3+} | \mathrm{Fe}^

In a cell, the following reactions take place

\begin{aligned} &\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \quad \mathrm{E}_{\mathrm{Fe}^{3+} | \mathrm{Fe}^{2+}=0.77 \mathrm{~V}}^{\circ} \\ &2 \mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-} \quad \mathrm{E}_{\mathrm{I}_{2} |\mathrm{I}^{-}}^{\circ}=0.54 \mathrm{~V} \end{aligned}

The standard electrode potential for the spontaneous reaction in the cell is \mathrm{x \times 10^{-2} \mathrm{~V}\ at\ 298 \mathrm{~K}} . The value of x is (Nearest Integer)

Option: 1

23


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The spontaneous reaction is

\mathrm{2 {Fe}^{3+} +2 I^{-} \longrightarrow 2 {Fe}^{2+}+I_{2} }

\begin{aligned} E^{0} &=E_{\mathrm{Fe}^{3+}\; |\mathrm{Fe}^{2+}}^{0} -E^0_\mathrm{{I_{2} |I^{-}}} \\ &=0.77-0.54 \\ &=0.23 \mathrm{~V} \end{aligned}

Thus, the electrode potential of the spontaneous reaction is \mathrm{23 \times 10^{-2} V}

Hence, the answer is 23.

Posted by

HARSH KANKARIA

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