Question

The $E^{\circ }$ in the given diagram is :-

• A. 0.5
• B. 0.63
• C. 0.7
• D. 0.8

Answer 1

Answer: (B)

$\underset{+5}{Cl{O}_3^{-}}\to \underset{+1}{Cl{O}^{-}}$	$n_1=4$	$E_1^{\circ }=0.50$	$\Delta G_1^{\circ }$
$\underset{+1}{Cl{O}^{-}}\to \frac{1}{2}\underset{0}{C{l}_2}$	$n_2=1$	$E_2^{\circ }=0.40$	$\Delta G_2^{\circ }$
$\frac{1}{2}\underset{0}{C{l}_2}\to \underset{-1}{C{l}^{-}}$	$n_3=1$	$E_3^{\circ }=1.36$	$\Delta G_3^{\circ }$
$\underset{+5}{Cl{O}_3^{-}}\to \underset{-1}{C{l}^{-}}$	$n=6$	$E^{\circ }=0$	$\Delta G_4^{\circ }=$ ?

$\Delta G^{\circ }=\Delta G_1^0+\Delta G_2^0+\Delta G_3^0$
$n{E}^{\circ }=n_1{E}_1^0+n_2{E}_2^0+n_3{E}_3^0$
$E^o=\frac{4\times 0.50+1\times 0.40+1\times 1.36}{6}$
$\Rightarrow 0.63volt$