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2 years ago

Idenify each statement below as either oxidation or reduction.

Chemistry

1 answer:

Alenkasestr [34]2 years ago

4 0

Answer:

1 = oxidation

2 = reduction

Explanation:

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

2I- ----> I₂+ 2e⁻

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

F + e⁻ ----> F⁻

Consider the following reactions.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

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3 years ago

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Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

$Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}$

Now, the two half-reactions can be written as follows.

Reduction half-reaction: $Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}$

This will be balanced as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$ ... (1)

Oxidation half-reaction: $Cl^{-} \rightarrow Cl_{2} + 1e^{-}$

This will be balanced as follows.

$6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-}$ ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O$

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

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2 years ago