Question

$A{s}_2{S}_3+N{O}_3^{⊝}+H_{2}O\to As{O}_4^{3-}+S{O}_4^{2-}+NO+H^{⨁}$
The number of electrons involved in the oxidation reaction is:

• A. $22$
• B. $24$
• C. $26$
• D. $28$

Answer 1

The correct option is D $28$

The balance reaction given below shows that the number of electrons involved is $28$.
$\stackrel{+3}{\underset{2x=6}{A{s}_2}}\to \underset{\underset{2x=10}{2x-16=-6}}{2As{O}_4^{3-}+4e^{-}}$
$\underset{3x=-6}{S_3^{-6}}\to \underset{\underset{3x=18}{3x-24=-6}}{3S{O}_4^{2-}+24e^{-}}$
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$A{s}_2{S}_3\to 2As{O}_4^{3-}+3S{O}_4^{2-}+28e^{-}$
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