For the reaction 4B_{(s)} + 3O_{2(g)} rightar | vedclass

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(D) Step $1$: Calculate $\Delta_rH^o$ for the reaction.$\Delta_rH^o = 2 	imes \Delta_fH^o(B_2O_3) - [4 	imes \Delta_fH^o(B) + 3 	imes \Delta_fH^o(O

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$150.3$

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(D) Step $1$: Calculate $\Delta_rH^o$ for the reaction.$\Delta_rH^o = 2 	imes \Delta_fH^o(B_2O_3) - [4 	imes \Delta_fH^o(B) + 3 	imes \Delta_fH^o(O_2)]$$\Delta_rH^o = 2 	imes (-840) - 0 = -1680 \ kJ/mol = -1680000 \ J/mol$.Step $2$: Calculate $\Delta_rG^o$ using $E^o_{cell}$.For the reaction,$n = 12$ (since $B^0 ightarrow B^{+3}$,$4 	imes 3 = 12$ electrons).$\Delta_rG^o = -nFE^o_{cell} = -12 	imes 96500 	imes 1.433 = -1659534 \ J/mol$.Step $3$: Calculate $\Delta_rS^o$ using $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$.$-1659534 = -1680000 - 298 	imes \Delta_rS^o$$298 	imes \Delta_rS^o = -1680000 + 1659534 = -20466$$\Delta_rS^o = -20466 / 298 = -68.677 \ J/K \ mol$.Step $4$: Calculate $S_m^o(O_2)$ using $\Delta_rS^o = \sum S_m^o(products) - \sum S_m^o(reactants)$.$-68.677 = [2 	imes S_m^o(B_2O_3)] - [4 	imes S_m^o(B) + 3 	imes S_m^o(O_2)]$$-68.677 = [2 	imes 280] - [4 	imes 10 + 3 	imes S_m^o(O_2)]$$-68.677 = 560 - 40 - 3 	imes S_m^o(O_2)$$-68.677 = 520 - 3 	imes S_m^o(O_2)$$3 	imes S_m^o(O_2) = 520 + 68.677 = 588.677$$S_m^o(O_2) = 588.677 / 3 = 196.22 \ J/K \ mol \approx 196.2 \ J/K \ mol$.Note: The provided options seem to be scaled by a factor of $1000$ or contain a typo. Based on standard values,$196.2 \ J/K \ mol$ is the correct magnitude.

List-$I$	List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$	$I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$	$II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$	$III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$	$IV$. Charging of secondary battery

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Based on the following information,arrange four metals,$A$,$B$,$C$,and $D$,in order of increasing ability to act as reducing agents:
$(I)$ Only $C$ reacts with $1 \ M \ HCl$ to give $H_{2(g)}$.
$(II)$ When $A$ is added to a solution of the other metal salts,metallic $D$ is formed,but not $B$ or $C$.

For a spontaneous reaction,the $\Delta G$,equilibrium constant $K$,and $E_{Cell}^{o}$ will be respectively:

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