Question

For the given reaction :

$S_8\left(s\right)+12NaOH\left(aq\right)\to 4N{a}_{2}S\left(aq\right)+2N{a}_2{S}_2{O}_3\left(aq\right)+6H_{2}O\left(l\right)$
What will be the overall valence factor $\left(n_f\right)$ for $S_8$?

• A. 16
• B. 8
• C. 5
• D. 20

Answer 1

The correct option is B 8

This is a disproportionation reaction where Sulphur simulataneously gets oxidised and reduced.

Reaction :
${\stackrel{0}{S}}_8\left(s\right)+12NaOH\left(aq\right)\to 4N{a}_2\stackrel{-2}{S}\left(aq\right)+2N{a}_2{\stackrel{+2}{S}}_2{O}_3\left(aq\right)+6H_{2}O\left(l\right)$

Oxidation :
${\stackrel{0}{S}}_8\left(s\right)⟶2N{a}_2{\stackrel{+2}{S}}_2{O}_3$
$(n_f{)}_{\text{oxidation}}=|0-(+2)|\times 8(n_f{)}_{\text{oxidation}}=16$

Reduction:
${\stackrel{0}{S}}_8\left(s\right)⟶4N{a}_2\stackrel{-2}{S}\left(aq\right)$
$(n_f{)}_{\text{reduction}}=|0-(-2)|\times 8(n_f{)}_{\text{reduction}}=16$

$(n_f{)}_{\text{overall}}=\frac{(n_f{)}_{\text{oxidation}}\times (n_f{)}_{\text{reduction}}}{(n_f{)}_{\text{oxidation}}+(n_f{)}_{\text{reduction}}}$

$(n_f{)}_{\text{overall}}=\frac{16\times 16}{16+16}=\frac{256}{32}(n_f{)}_{\text{overall}}=8$