In an electrochemical reaction of lead, at standard temperature, if E^0( Pb^2+ / Pb)= m Volt and E^0( Pb^4+ / Pb)= n Volt, then the value of E^0( Pb^2+ / Pb^4+) is given by m-x n. The value of x is ___________. (Nearest integer)

Pb^2++2 e^- -> Pb Delta G_1^0=-2 FE_1^0 Pb^4++4 e^- -> Pb Delta G_2^0=-4 FE_2^0 Pb^2+ -> Pb^4++2 e^- Delta G_3^0=-2 FE_3^0 Delta G_3^0=Delta G_1^0-Delta G_2^0 -2 FE_3^0=2 F(2 n- m) E_3^0= m-2 n= m- xnHence x=2

In an electrochemical reaction of lead at standard temp... | (Chemistry)

$\mathrm{Pb}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb} \quad \Delta \mathrm{G}_1^0=-2 \mathrm{FE}_1^0 \\$

$\mathrm{~Pb}^{4+}+4 \mathrm{e}^{-} \rightarrow \mathrm{Pb} \quad \Delta \mathrm{G}_2^0=-4 \mathrm{FE}_2^0 \\$

$\mathrm{~Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-} \quad \Delta \mathrm{G}_3^0=-2 \mathrm{FE}_3^0 \\$

$\Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0-\Delta \mathrm{G}_2^0 \\$

$-2 \mathrm{FE}_3^0=2 \mathrm{~F}(2 \mathrm{n}-\mathrm{m}) \\$

$\mathrm{E}_3^0=\mathrm{m}-2 \mathrm{n}=\mathrm{m}-\mathrm{xn}$

Hence $\mathrm{x}=2$

Explanation

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