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

Open in App

Solution

$P_{4} + O H^{-} ⟶ P H_{3} + H P O_{2}^{-} o - 3 + 2$
$P_{4}$ : reduction and $P_{4}$ : oxidation.

$2 P_{4} + 3 P_{4} + 4 (3 O H^{-}) + 3 (H_{2} O) ⟶ 8 P H_{3} + 12 H P O_{3}^{-}$
$5 P_{4} + 4 (3 O H^{-} + 3 H_{2} O) ⟶ 8 P H_{3} + 12 H P O_{2}^{-}$
$5 P_{4} + 12 O H^{-} + 12 H_{2} O ⟶ 8 P H_{3} + 12 H P O_{2}^{-}$

Suggest Corrections

Similar questions

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

C l_{2} O_{7} (g) + H_{2} O_{2} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g) + H^{+}

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

$N_{2} H_{4 (I)} + C I O_{3 (a q)}^{-} ⟶ N O_{(g)} + C I_{(g)}^{-}$

$C l_{2} O_{7 (g)} + H_{2} O_{2 (a q)} ⟶ C I O_{2 (a q)}^{-} + O_{2 (g)} + H_{(a q)}^{+}$

Q. Balance the following equations in acidic medium by both oxidation number and ion electron methods and identify the oxidants and the reductants :

M n O_{4}^{-} (a q) + C_{2} H_{2} O_{4} (a q) \to M n^{2 +} (a q) + C O_{2} (g) + H_{2} O (l)

Q. Balance the following equation by ion electron method.

P_{4} + O H^{-} + H_{2} O \to H_{2} P O_{2}^{-} + P H_{3}

Q. Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction.

{H C H O}_{(l)} + 2 [A g (N H_{3})_{2}]_{(a q)}^{+} + 3 O H_{(a q)}^{-} \to 2 A g_{(s)} + H C O O_{(a q)}^{-} + 4 {N H_{3}}_{(a q)} + 2 {H_{2} O}_{(l)}

Join BYJU'S Learning Program

Balance the following equation in a basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.P4(s)+OH−(aq)→PH3(g)+HPO−2(aq).

P4+OH−⟶PH3+HPO−2o−3+2P4: reduction and P4: oxidation.2P4+3P4+4(3OH−)+3(H2O)⟶8PH3+12HPO−35P4+4(3OH−+3H2O)⟶8PH3+12HPO−25P4+12OH−+12H2O⟶8PH3+12HPO−2

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