The rusting of iron takes place as follows. C | vedclass

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(A) The net reaction is the sum of the oxidation and reduction half-reactions:$Fe_{(s)} \longrightarrow Fe^{2+} + 2e^- ; E^o_{ox} = +0.44 \ V$$2H^{+}

Vedclass · Accepted Answer

$-322$

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(A) The net reaction is the sum of the oxidation and reduction half-reactions:$Fe_{(s)} \longrightarrow Fe^{2+} + 2e^- ; E^o_{ox} = +0.44 \ V$$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o_{red} = +1.23 \ V$Net reaction: $Fe_{(s)} + 2H^{+} + \frac{1}{2}O_2 \longrightarrow Fe^{2+} + H_2O_{(l)}$$E^o_{cell} = E^o_{red} + E^o_{ox} = 1.23 \ V + 0.44 \ V = 1.67 \ V$Using the formula $\Delta G^o = -nFE^o_{cell}$,where $n = 2$ and $F = 96500 \ C \ mol^{-1}$:$\Delta G^o = -2 	imes 96500 	imes 1.67 \ J \ mol^{-1}$$\Delta G^o = -322310 \ J \ mol^{-1} = -322.31 \ kJ \ mol^{-1}$Rounding to the nearest integer,we get $-322 \ kJ \ mol^{-1}$.

List-$I$ (Conversion)	List-$II$ (Number of Faraday required)
$A$. $1 \ mol$ of $H_2O$ to $O_2$	$I$. $3 \ F$
$B$. $1 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$	$II$. $2 \ F$
$C$. $1.5 \ mol$ of $Ca$ from molten $CaCl_2$	$III$. $1 \ F$
$D$. $1 \ mol$ of $FeO$ to $Fe_2O_3$	$IV$. $5 \ F$

The rusting of iron takes place as follows. Calculate $\Delta G^o$ for the net process in $kJ \ mol^{-1}$.
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$
$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$

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The rusting of iron takes place as follows. Calculate $\Delta G^o$ for the net process in $kJ \ mol^{-1}$.$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$