Consider the following reaction- Mx-MnO-4- MO-3-Mn2-If 1 mol of MnO-4 oxidises 1-67 mol of Mx- to MO-3- then the value of x in the reaction is -

The correct option is A 2 Given the following reaction: M^x++MnO^⊝_4→ MO^⊝_3+Mn^2+ The half cell reactions are as follows: MnO^⊝_4+8H^⊕+5e^ → Mn^2 ...

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Standard XII

Chemistry

Valency and Oxidation State

Consider the ...

Question

Consider the following reaction:
$M^{x +} + M n O_{4}^{⊝} \to M O_{3}^{⊝} + M n^{2 +}$
If $1$ mol of $M n O_{4}^{⊝}$ oxidises $1.67$ mol of $M^{x +}$ to $M O_{3}^{⊝}$ , then the value of $x$ in the reaction is :

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Solution

The correct option is A $2$
Given the following reaction:
$M^{x +} + M n O_{4}^{⊝} \to M O_{3}^{⊝} + M n^{2 +}$
The half cell reactions are as follows:
$M n O_{4}^{⊝} + 8 H^{⨁} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
$M^{x +} + 3 H_{2} O \to M O_{3}^{⊝} + 6 H^{⨁} + (5 - x) e^{-}$
mEq. of $M n O_{4}^{2 -} \equiv$ mEq. of $M^{x +}$
$1 \times 5 \equiv [1.67 \times (5 - x)] \Rightarrow 5 - x = 3 \Rightarrow x = 2$
Therefore, $x$ is $2$ .

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Similar questions

Q. For the reaction,

M^{x +} + M n O_{4}^{-} \to M O_{3}^{-} + M n^{2 +}

If one mole of

M n O_{4}^{-}

oxidises 1.67 moles of

M^{x +}

to oxidation state where its magnet moment is 1.73. Find x. ___

Q. For a reaction

M^{x +} + M n O_{4}^{-} \to M O_{3}^{-} + M n^{2 +} + \frac{1}{2} O_{2}

If 1 mole of

M n O_{4}^{-}

oxidises 1.66 moles of

M^{x +}

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Q. Given a reaction:

M^{x +} + M n O_{4}^{-} ⟶ M O_{3}^{-} + {M n}^{2 +} + \frac{1}{2} O_{2}

1

mol of

M n O_{4}^{-}

oxidises

1.67

mol of

M^{x +}

M O_{3}^{-}

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in this reaction is:

Q. For the following reaction:

M^{x +} + M n O_{4}^{⊖} ⟶ M O_{3}^{⊖} + {M n}^{2 +} + (\frac{1}{2}) O_{2}

1

mol of

M n O_{4}^{⊖}

oxidises

1.67

mol of

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x

in the reaction is :

M^{x +} + M n O_{4}^{- 1} \to M O_{3}^{- 1} + M n^{+ 2}

If 1 mol of

M n O_{4}^{- 1}

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Consider the following reaction:Mx++MnO⊝4→MO⊝3+Mn2+If 1 mol of MnO⊝4 oxidises 1.67 mol of Mx+ to MO⊝3, then the value of x in the reaction is :

The correct option is A 2Given the following reaction:Mx++MnO⊝4→MO⊝3+Mn2+The half cell reactions are as follows:MnO⊝4+8H⨁+5e−→Mn2++4H2OMx++3H2O→MO⊝3+6H⨁+(5−x)e−mEq. of MnO2−4≡ mEq. of Mx+1×5≡[1.67×(5−x)]⇒5−x=3⇒x=2Therefore, x is 2.

Consider the following reaction:
$M^{x +} + M n O_{4}^{⊝} \to M O_{3}^{⊝} + M n^{2 +}$
If $1$ mol of $M n O_{4}^{⊝}$ oxidises $1.67$ mol of $M^{x +}$ to $M O_{3}^{⊝}$ , then the value of $x$ in the reaction is :