The value of n in the following process: B O 3 n − + 4e → HB O 2 2 − is

Balance with H + and H 2 O : B O 3 n − + a H + + 4e → HB O 2 2 − + b H 2 O . Charge: -n + a - 4 = -2 ⇒ a = n + 2. O: 3 = 2 + b ⇒ b = 1. H: a = 1 + 2(1) = 3, so n + 2 = 3 ⇒ n = 1. Ans: 1.

Start Free Mock LoginAlready registered? Login here

Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsProblem involving Equivalent ConceptEasy2 minQB

ChemistryEasynumerical

The value of $n$ in the following process: $B O_{3}^{n -}$ + 4e $\to$ $HB O_{2}^{2 -}$ is

Answer:

1.00

Solution:

Balance with $H^{+}$ and $H_{2} O$ : $B O_{3}^{n -}$ + $a H^{+}$ + 4e $\to$ $HB O_{2}^{2 -}$ + $b H_{2} O$ . Charge: -n + a - 4 = -2 $\Rightarrow$ a = $n$ + 2. O: 3 = 2 + b $\Rightarrow$ b = 1. H: a = 1 + 2(1) = 3, so $n$ + 2 = 3 $\Rightarrow$ $n$ = 1. Ans: 1.

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Problem involving Equivalent Concept

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...