If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Question

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

in progress 0
Thành Đạt 6 months 2021-07-11T21:07:04+00:00 1 Answers 10 views 0

Answers ( )

    0
    2021-07-11T21:08:46+00:00

    Answer:

    1.88 A

    Explanation:

    Let’s consider the reduction of copper in an electrolytic cell.

    Cu²⁺ + 2 e⁻ ⇒ Cu

    We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

    • The molar mass of Cu is 63.55 g/mol.
    • 1 mole of Cu is deposited when 2 moles of electrons circulate.
    • 1 mole of electrons has a charge of 96486 C (Faraday’s constant).

    The charge used is:

    12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron}  = 3.73 \times 10^{4} C

    We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

    5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

    The current used is:

    I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )