SATHEE: Transition and InnerTransition Elements

Transition and InnerTransition Elements - Result Question 96

####23. Consider the following reaction,

(2013 Main) $x M n O_{4}^{-} + y C_{2} O_{4}^{2 -} + z H^{+} ⟶ x M n^{2 +} + 2 y C O_{2} + \frac{z}{2} H_{2} O$ The values of $x, y$ and $z$ in the reaction are, respectively

(a) 5, 2 and 16

(b) 2,5 and 8

(d) 5,2 and 8

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Solution:

The half equations of the reaction are

$\begin{aligned} M n O_{4}^{-} ⟶ M n^{2 +} & C_{2} O_{4}^{2 -} ⟶ C O_{2} \end{aligned}$

The balanced half equations are

$\begin{aligned} M n O_{4}^{-} + 8 H^{+} + 5 e^{-} & ⟶ M n^{2 +} + 4 H_{2} O C_{2} O_{4}^{2 -} & ⟶ 2 C O_{2} + 2 e^{-} \end{aligned}$

On equating number of electrons, we get

$\begin{aligned} 2 M n O_{4}^{-} + 16 H^{+} + 10 e^{-} & ⟶ 2 M n^{2 +} + 8 H_{2} O 5 C_{2} O_{4}^{2 -} & ⟶ 10 C O_{2} + 10 e^{-} \end{aligned}$

On adding both the equations, we get

$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{-} + 16 H^{+} ⟶ 2 M n^{2 +} + 2 \times 5 C O_{2} + \frac{16}{2} H_{2} O$

Thus $x, y$ and $z$ are 2, 5 and 16 respectively.

Transition and InnerTransition Elements - Result Question 96

Solution:

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Transition and InnerTransition Elements - Result Question 96

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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