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It took 2.30 minutes using a current of 3.00 A to plate out all the copper from 0.300 L of a solution containing Cu2 . What was the original concentration of Cu2

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dsdrajlin

Answer:

7.16 × 10⁻³ M

Explanation:

Let's consider the reduction reaction of copper during the electroplating.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We can calculate the moles of Cu²⁺ present in the solution using the following relations.

  • 1 A = 1 C/s.
  • 1 min = 60 s.
  • 1 mole of electrons has a charge of 96486 C (Faraday's constant).
  • 1 mole of Cu²⁺ is reduced when 2 moles of electrons are gained.

The moles of Cu²⁺ reduced are:

[tex]2.30 min \times \frac{60s}{1min} \times \frac{3.00C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molCu^{2+} }{2mole^{-} } = 2.15 \times 10^{-3} molCu^{2+}[/tex]

[tex]2.15 \times 10^{-3} moles[/tex] of Cu²⁺ are in 0.300 L of solution.

[Cu²⁺] = 2.15 × 10⁻³ mol/0.300 L = 7.16 × 10⁻³ M

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