Question

Balance the following equations in basic medium by ion-elecron method and oxidation number methods and identify the oxidizing agent and reducing agent.

${\sf{1) \: P_{4}(s) \: + \: OH^{-}(aq) \: {\rightarrow} \: PH_{3}(g) \: + \: H_{2}PO_{2}^{-}}}$

${\sf{2) \: N_{2}H_{4}(l) \: + \: ClO_{3}^{-}(aq) {\rightarrow} \: NO(g) \: + \: Cl^{-}(g)}}$​

Answer 1

Ion-Electron Method:-

Given:-

${\sf{1) \: P_{4}(s) \: + \: OH^{-}(aq) \: {\rightarrow} \: PH_{3}(g) \: + \: H_{2}PO_{2}^{-}}}$

${\sf{2) \: N_{2}H_{4}(l) \: + \: ClO_{3}^{-}(aq) {\rightarrow} \: NO(g) \: + \: Cl^{-}(g)}}$

To Find:

Balanced equations

Solution:

Let us balance 1st equation

${\sf{1) \: P_{4}(s) \: + \: OH^{-}(aq) \: {\rightarrow} \: PH_{3}(g) \: + \: H_{2}PO_{2}^{-}}}$

$P_{4}(s)$ is oxidized and formed $H_{2}PO_{2}^{-}$

$P_{4}(s)$ is reduced and formed $PH_{3}(g)$

Hence,

We can split the equations as

OXIDATION HALF REACTION:

${\sf{1) \: P_{4}(s) \: {\rightarrow} \: H_{2}PO_{2}^{-}}}$

RECUCED HALF REACTION:

${\sf{1) \: P_{4}(s) \: {\rightarrow} \: PH_{3}(g)}}$

Now, balance the OHR and RHR

OXIDATION HALF REACTION:

${\sf{ \: P_{4}(s) \: {\rightarrow} \: H_{2}PO_{2}^{-}}}$

• Balance P atoms on the right side..by writing Stoichimetric coefficient 4

${\sf{\: P_{4}(s) \: {\rightarrow} \: \bold{4}H_{2}PO_{2}^{-}}}$

• Now, add ${OH}^{-}(aq)$ on left side. Since, there is a shortage of oxygen atoms. Now, place a suitable coefficient before it.

${\sf{ \:\bold{8OH^{-}(aq)}+ P_{4}(s) \: {\rightarrow} \: \bold{4}H_{2}PO_{2}^{-} }}$

• Here, Hydrogen atoms are also balanced. Hence There is no requirement of adding water molecule

• But, Charge should be balanced! . So, add 4 electorns on Right side.

${\sf{\:\bold{8OH^{-}(aq)}+ P_{4}(s) \: {\rightarrow} \: \bold{4}H_{2}PO_{2}^{-} + \bold{4{e}^{-}}}}-----(1)$

RECUCED HALF REACTION:

${\sf{ \: P_{4}(s) \: {\rightarrow} \: PH_{3}(g)}}$

• Balance P atoms on right side by adding coefficient 4

${\sf{\: P_{4}(s) \: {\rightarrow} \: \bold{4}PH_{3}(g)}}$

• Now, balance Hydrogen atoms by Adding Water molecule and using particular coefficient on Left side.

${\sf{ \:\bold{12 H_2O }P_{4}(s) \: {\rightarrow} \: \bold{4}PH_{3}(g)}}$

• Now balance Oxygen atoms by adding Hydroxyl ion and using particular coefficient on right side

${\sf{ \:\bold{12 H_2O }P_{4}(s) \: {\rightarrow} \: \bold{4}PH_{3}(g) + \bold{12OH^{-}(aq) }}}$

• Balance the charge on by adding 12 electrons on left side

${\sf{ \:\bold{ 12{e}^{-}+ 12 H_2O }P_{4}(s) \: {\rightarrow} \: \bold{4}PH_{3}(g) + \bold{12OH^{-}(aq) } }}---(2)$

Now, make sure that electors in eq. (1) and (2) should be equal and add both equatios

eq(1) × 3 + eq(2)

Then, we will get,

${\sf{ \:12 H_2O+ 12OH^{-}(aq)+ 4 P_{4}(s) \: + \: OH^{-}(aq) \: {\rightarrow} \:4 PH_{3}(g) \: + \:12 H_{2}PO_{2}^{-}}}$

The above equation is balanced

Let us balance equation (2)

${\sf{2) \: N_{2}H_{4}(l) \: + \: ClO_{3}^{-}(aq) {\rightarrow} \: NO(g) \: + \: Cl^{-}(g)}}$

On splitting the reactions into OHR and RHR

OHR:

${\sf{\: N_{2}H_{4}(l) \: {\rightarrow} \: NO(g) \: }}$

• Balance N atoms by writing Coefficient 2 on right side

${\sf{\: N_{2}H_{4}(l) \: {\rightarrow} \: \bold{2}NO(g) \: }}$

• Balance, Oxygen atoms and Hydrogen atoms by adding Hydroxyl ion and water molecule respectively. Also, use particular coefficients

${\sf{\:\bold{ 8OH^{-}(aq)}+ N_{2}H_{4}(l) \: {\rightarrow} \: \bold{2}NO(g) \: + \bold{6H_2O} }}$

• Now, balance the charge by adding 8 electrons on right side

${\sf{\:\bold{ 8OH^{-}(aq)}+ N_{2}H_{4}(l) \: {\rightarrow} \: \bold{2}NO(g) \: + \bold{6H_2O} + \bold{8{e}^{-}}}}---(1)$

RHR:

${\sf{ \: ClO_{3}^{-}(aq) {\rightarrow} \: Cl^{-}(g)}}$

• Cl atoms are balanced. Now, balance Oxygen atoms and Hydrogen atoms on both side by adding Hydroxyl ions and water molecule respectively.

${\sf{ \: \bold{3H_2O}ClO_{3}^{-}(aq) {\rightarrow} \: Cl^{-}(g) + \bold{ 6OH^{-}(aq)}}}$

• Balance the charge by adding 6 electrons on left side

${\sf{ \:\bold{6{e}^{-1} }+ \bold{3H_2O}ClO_{3}^{-}(aq) {\rightarrow} \: Cl^{-}(g) + \bold{ 6OH^{-}(aq)}}}$

Eq. 3 × (1) + 4 × (2)

${\sf{ \: 3N_{2}H_{4}(l) \: + \: 4ClO_{3}^{-}(aq) {\rightarrow} \: 3NO(g) \: + \: 4Cl^{-}(g) + 6H_2O }}$

The above equation is balanced