For the following oxidation half reaction- 2 S 2 O 32- S 4 O 62-2 e -Reducing agent Calculate the equivalent weight of S 2 O 32- Given- M is the molecular weight of S 2 O 32- A- MB- 2 MC- 1-5 MD- 0-5 M

The correct option is A MThe given equation is, 2+2S_2O_3^2 →+2.5S_4O_6^2 + 2e^ formula used for the n factor calculation, n_f=(|O.S._Product O.S._Reactant|× ...

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Standard XII

Chemistry

Equivalent Mass

For the follo...

Question

For the following oxidation half reaction,
$2 S_{2} O_{3}^{2 -} (Reducing ~agent) \to S_{4} O_{6}^{2 -} + 2 e^{-}$
Calculate the equivalent weight of $S_{2} O_{3}^{2 -}$ .
(Given: $M$ is the molecular weight of $S_{2} O_{3}^{2 -}$ )

0.5 M

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M

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1.5 M

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2 M

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Solution

The correct option is B $M$
The given equation is,
$2 {+ 2 S}_{2} O_{3}^{2 -} \to {+ 2.5 S}_{4} O_{6}^{2 -} + 2 e^{-}$
formula used for the n-factor calculation,
$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atoms$
$n_{f} = | 2 - 2.5 | \times 2 = 1$
we know, $Equivalent weight = \frac{Molecular weight}{n_{f}}$
$∴ Equivalent weight of S_{2} O_{3}^{2 -} = \frac{M}{1} = M$

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Equivalent Mass

Standard XII Chemistry

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For the following oxidation half reaction, 2S2O2−3(Reducing ~agent)→S4O2−6+2e− Calculate the equivalent weight of S2O2−3. (Given: M is the molecular weight of S2O2−3)

The correct option is B MThe given equation is, 2+2S2O2−3→+2.5S4O2−6+2e− formula used for the n-factor calculation, nf=(|O.S.Product−O.S.Reactant|×number of atoms nf=|2−2.5|×2=1 we know, Equivalent weight =Molecular weightnf ∴Equivalent weight of S2O2−3=M1=M

For the following oxidation half reaction,
$2 S_{2} O_{3}^{2 -} (Reducing ~agent) \to S_{4} O_{6}^{2 -} + 2 e^{-}$
Calculate the equivalent weight of $S_{2} O_{3}^{2 -}$ .
(Given: $M$ is the molecular weight of $S_{2} O_{3}^{2 -}$ )