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beks73 [17]
3 years ago
13

Write a balanced half-reaction describing the reduction of gaseous dioxygen to aqueous oxide anions.

Chemistry
1 answer:
maksim [4K]3 years ago
8 0

<u>Answer:</u> The balanced half reaction is written below.

<u>Explanation:</u>

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

When oxygen gas is reduced to oxide ions, the number of electron transferred are 2

The chemical equation for the reduction of oxygen gas to oxide ions follows:

O_2+2e^-\rightarrow 2O^{2-}

Hence, the balanced half reaction is written above.

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Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

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2 moles of SO_2 require = 1 mole of O_2

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Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

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Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

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