Name: Hydrogen Electrode
Uploaded: 2019-07-19
Duration: 3 min 7 s
Description: Every unit of measurement has a standard measurement associated with it. The same is with the electrode potential. Let's learn what the standard measurement...

Question

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

P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H P O_{2}^{-} (a q)

Accepted Answer

$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H P O_{2}^{-} (a q)$
The oxidation number (O.N.) of $P$ decreases from $0$ in $P_{4}$ to $- 3$ in $P H_{3}$ and increases from $0$ in $P_{4}$ to $+ 2$ in $H P O_{2}^{-}$ . Hence, $P_{4}$ acts both as an oxidizing agent and a reducing agent in this reaction.

Ion–electron method:
The oxidation half equation is:
$P_{4} (s) \to H P O_{2}^{-} (a q)$
The $P$ atom is balanced as:
$P_{4} \to 4 H P O_{2}^{-} (a q)$
The $O$ atom is balanced by adding $8 H_{2} O$ molecules:
$P_{4} + 8 H_{2} O \to 4 H P O_{2}^{-} (a q)$
The $H$ atom is balanced by adding $12 H^{+}$ ions:
$P_{4} + 8 H_{2} O \to 4 H P O_{2}^{-} (a q) + 12 H^{+}$
The charge is balanced by adding 8 $e^{-}$ as:
$P_{4} + 8 H_{2} O \to 4 H P O_{2}^{-} (a q) + 12 H^{+} + 8 e^{-}$ …(i)
The reduction half equation is:
$P_{4} (s) \to P H_{3} (g)$
The $P$ atom is balanced as:
$P_{4} \to 4 P H_{3} (g)$
The $H$ is balanced by adding $12 H^{+}$ as:
$P_{4} + 12 H^{+} \to 4 P H_{3} (g)$
The charge is balanced by adding $12 e^{-}$ as:
$P_{4} + 12 H^{+} + 12 e^{-} \to 4 P H_{3} (g)$ …(ii)
By multiplying equation (i) with $3$ and (ii) with $2$ and then adding them, the balanced chemical equation can be obtained as:
$5 P_{4} (s) + 24 H_{2} O \to 12 H P O_{2}^{-} + 8 P H_{3} (g) + 12 H^{+}$
As, the medium is basic, add $12 {O H}^{-}$ both sides as:
$5 P_{4} (s) + 12 H_{2} O + 12 O H^{-} \to 12 H P O_{2}^{-} + 8 P H_{3} (g)$
This is the balanced equation by ion-electron method.

Oxidation number method:
Let, total number of $P$ reduced $= x$
$∴$ Total number of $P$ oxidised $= 4 - x$
$P_{4} (s) + O H^{-} \to x P H_{3} (g) + (4 - x) H P O_{2}^{-}$ …(i)
Total decrease in O.N. of $P = x \times 3 = 3 x$
Total increase in O.N. of $P = (4 - x) \times 2 = 8 - 2 x$
$∵ 3 x = (8 - 2 x) \Rightarrow x = 8 / 5$

Then, put the value of x in the equation(i)
$5 P_{4} (s) + 5 O H^{-} \to 8 P H_{3} (g) + 12 H P O_{2}^{-}$
Since, reaction occurs in basic medium, the charge is balanced by adding $7 O H^{-}$ on LHS as:
$5 P_{4} (s) + 12 O H^{-} \to 8 P H_{3} (g) + 12 H P O_{2}^{-}$
The $O$ atoms are balanced by adding $12 H_{2} O$ as:
$5 P_{4} (s) + 12 H_{2} O + 12 O H^{-} \to + 12 H P O_{2}^{-} + 8 P H_{3} (g)$
This is the required balanced equation.

Balance the following equation in the basic medium by ion-electron method and oxidation number methods and identify the oxidizing agent and the reducing agent.
$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H P O_{2}^{-} (a q)$

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !

Balance the following equation in the basic medium by ion-electron method and oxidation number methods and identify the oxidizing agent and the reducing agent. P4s+OH-aq→PH3g+HPO2-(aq)

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !

Balance the following equation in the basic medium by ion-electron method and oxidation number methods and identify the oxidizing agent and the reducing agent.
$P_{4} (s) + O H^{-} (a q) \to P H_{3} (g) + H P O_{2}^{-} (a q)$