Question

Balance the following equation in acidic medium by both oxidation number and ion electron method and identify the oxidants and the reductants: $H_{2}S(aq)+Cl_{2}(g) \longrightarrow S(s)+Cl^{-}(aq)$

Answer 1

When Hydrogen sulphide reacts with chlorine gas to give solid sulphur and HCl

By balancing the equation we get,  

i) Oxidation number method:

The balanced equation is as follows,

${ H }_{ 2 }S(aq)\quad +\quad { Cl }_{ 2 }(g)\quad \rightarrow \quad S(s)\quad +\quad 2HCl(aq)$

Oxidation: $H\quad +\quad 2{ S }^{ -2 }\quad \rightarrow \quad S$

Reduction: ${ Cl }^{ 0 }\quad \rightarrow \quad { H }^{ +1 }{ Cl }^{ -1 }$

Where +2 state of Sulphur in the reactant gets oxidized to zero state in the product and Zero state chlorine in the reactants gets reduced to ${ Cl }^{ - }$ in the product

ii) Ion electron method:

${ H }_{ 2 }S(aq)\quad +\quad { Cl }_{ 2 }(g)\quad \rightarrow \quad S(s)\quad +\quad 2HCl(aq)$

a) $O: { H }_{ 2 }S \rightarrow S$

$R: Cl \rightarrow HCl$

b) $O: { H }_{ 2 }S \rightarrow S$

$R: Cl \rightarrow HCl$

c) $O: { H }_{ 2 }S \rightarrow S + 2{ H }^{ + }$

$R: Cl + { H }^{ + } \rightarrow HCl$

d) $O: { H }_{ 2 }S \rightarrow S + 2{ H }^{ + } + 2{ e }^{ - }$

$R: Cl + { H }^{ + } + { e }^{ - } \rightarrow HCl$

e) $O: { H }_{ 2 }S \rightarrow S + 2{ H }^{ + } + 2{ e }^{ - }$

$R: Cl + { H }^{ + } + { e }^{ - } \rightarrow HCl$

f) $O: { H }_{ 2 }S \rightarrow S + 2{ H }^{ + } + 2{ e }^{ - }$

$R: 2Cl + 2{ H }^{ + } + 2{ e }^{ - } \rightarrow 2HCl$

g) ${ H }_{ 2 }S + 2Cl + 2{ H }^{ + } + 2{ e }^{ - } \rightarrow S + 2HCl + 2{ H }^{ + } + 2{ e }^{ - }$

This the final reaction where L.H.s is equal to R.H.S

h) ${ H }_{ 2 }S + 2Cl \rightarrow S + 2HCl$

i) ${ H }_{ 2 }S(aq)\quad +\quad { Cl }_{ 2 }(g)\quad \rightarrow\quad S(s)\quad +\quad 2HCl(aq)$

This is a best example for redox reaction. Here, chlorine is the oxidizing agent which gets reduced itself to HCl and ${ H }_{ 2 }S$ is the reducing agent which gets oxidized to sulphur.