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Electrochemistry question

2023 · 31 Jan · Shift 1 · Q17

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Electrochemistry
2023 · 31 Jan · Shift 1 · Q17

Electrochemistry question

2023 · 31 Jan · Shift 1 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1

The logarithm of equilibrium constant for the reaction

\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-}

is

\underline{\hspace{2cm}}

(Nearest integer) Given :

\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V} \mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V} \begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned}

Numerical answer

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