For a cell involving one electron E_ cell ^ =0.59 V at 298 ;K, the equilibrium constant for the cell reaction is Given that . 2.303 RT F =0.059 V at T=298 K ]

For a cell involving one electron E_ cell ^ =0.59 V at 298 ;K, th…

MCQ

For a cell involving one electron $E_{cell }^{\ominus} =0.59 \mathrm{V}$ at $298 \;\mathrm{K}$, the equilibrium constant for the cell reaction is

Given that $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V} \text { at } \mathrm{T}=298 \mathrm{K}\right]$

✓

Correct option: C.

$1.0 \times 10^{10}$

c
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log _{10} \mathrm{Q}$

at equllbrium $\mathrm{E}_{\mathrm{cell}}=0, \mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$

$0=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0591}{1} \log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{E}_{\text {cell }}^{\circ}=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq}}$

$0.59=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq} .}$

$+10=\log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{K}_{\mathrm{eq}}=10^{+10}$

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Given $: \frac{2.303 RT }{ F }=0.06 V$

$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$

$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$

$E ^{\circ}=0.65\,V$

List $I$ (Conversion)	List $II$ (Number of Faraday required)
$A$. $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$	$l$. $3 \mathrm{~F}$
$B$. $1 \mathrm{~mol}$ of $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$	$II$. $2 F$
$C$. $1.5 \mathrm{~mol}$ of $\mathrm{Ca}$ from molten $\mathrm{CaCl}_2$	$III$. $1F$
$D$. $1 \mathrm{~mol}$ of $\mathrm{FeO}$ to $\mathrm{Fe}_2 \mathrm{O}_3$	$IV$. $5 \mathrm{~F}$