Question

Balance the redox equation in acidic medium.

Answer 1

Answer

$\sf\tiny{3S^{-2} + 2MnO_4^- + 8H^+\rightarrow 3S + 2MnO_2 + 4H_2O}$

$\rule{200}2$

Explanation

Given reaction is redox reaction. In acidic medium, to balance redox reaction, use H2O to balance oxygen and finally H+ to balance hydrogen.

$\sf S^{2-} + MnO_4^{-} \Rightarrow S + MnO_2$

Using half reaction method of balancing -

Reduction -

In MnO4-, oxidation number of Mn is +7 and in MnO2, it is +4.

Hence, number of electrons to be added is 3 electrons on LHS.

$\sf 3e^- + MnO_4^- \Rightarrow MnO_2$

Next step is to balance oxygen by adding water.

$\sf 3e^- + MnO_4^- \Rightarrow MnO_2 + 2H_2O$

Now, balance Hydrogen by adding H+ ions.

$\sf 3e^- + MnO_4^- + 4H^+ \rightarrow MnO_2 + 2H_2O$

Now, balance oxidation half

$\sf S^{2-} \Rightarrow S$

The oxidation number is changing from -2 to 0. Add 2 electrons in RHS.

$\sf S^{2-} \Rightarrow S + 2e^-$

Multiply the reduction half with 2 and oxidation half with 3 to cancel out electrons.

Oxidation half

$\sf 3S^{2-} \Rightarrow 3S + 6e^-$

(after multiplying with 3)

Reduction half

$\sf 6e^- + 2MnO_4^- + 8H^+ \rightarrow 2MnO_2 + 4H_2O$

Add the oxidation half and reduction half

$\sf \tiny{6e^- + 2MnO_4^- + 8H^+ 3S^{-2} \rightarrow 2MnO_2 + 4H_2O + 3S + 6e^-}$

Cancel electrons

$\sf \tiny{3S^{-2} + 2MnO_4^- + 8H^+\rightarrow 3S + 2MnO_2 + 4H_2O}$

Verification

★Check for charge :

LHS charge = 3(-2) + 2(-1) + 8

= -6 - 2 + 8

= 0

RHS charge = 0

Charge balanced.

$\rule{25}1$

★ Check for elements

LHS Mn = 2

RHS Mn = 2

Manganese Balanced.

LHS, S = 3

RHS, S = 3

Sulphur balanced.

LHS H = 8 = RHS

Hydrogen balanced.

LHS O = 8 = RHS

Oxygen balanced.