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jolli1 [7]
3 years ago
15

Iodine, I2, has many uses, including the production of dyes, antiseptics, photographic film, pharmaceuticals, and medicinal soap

s. It forms when chlorine, Cl2, reacts with iodide ions in a sodium iodide solution. Which of the following half-reactions for this oxidation‑reduction reaction describes the oxidation, and which one describes the reduction? Cl2 + 2e‒ → 2Cl‒ 2I‒ → I2 + 2e‒
Chemistry
1 answer:
svetoff [14.1K]3 years ago
4 0

Answer:

  • <em><u>2I ⁻ → I₂ + 2e⁻</u></em> describes the <em><u>oxidation.</u></em>

  • <u><em>Cl₂ + 2e⁻ → 2Cl ⁻</em></u> describes the <u><em>reduction</em></u>.

Explanation:

<em>Oxidation-reduction reaction</em> is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.

The <em>oxidation</em> half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.

In this case, the process that shows the loss of electrons is:

  • 2I⁻ → I₂ + 2e⁻

That reaction shows:

  • Two I⁻ ions lose two electrons (one each) to be oxidized to I₂.
  • The change in the oxidation number is from -1 to 0.
  • Hence this half-reaction is the oxidation reaction.

On the other hand, the <em>reduction</em> half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.

In this case, the process that shows the gain of electrons is:

  • Cl₂ + 2e⁻ → 2Cl⁻

That reaction shows:

  • Two Cl atoms gain two electrons (one each) to be reduced to Cl⁻.
  • The change in the oxidation number is from 0 to - 1.
  • Hence, this half-reaction is the reduction reaction.

<u>Summary:</u>

  • <em>2I ⁻ → I₂</em> + 2e⁻ describes the oxidation.

  • <em>Cl₂ + 2e⁻ → 2Cl ⁻</em> describes the reduction.
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Theoretical yield = 2.397

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percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

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mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

#SPJ4

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