Question

You perform an electrochemical reaction in which 0.600 mol of Cu are reduced to solid Cu. How many coulombs of charge are transferred?How many electrons are in this amount of charge?

Answer 1

Answer:

$7.22 \times 10^{23}e^{-}$

$1.16 \times 10^{5}C$

Explanation:

Step 1: Write the reaction for the electrochemical reduction of copper.

Cu²⁺(aq) + 2 e⁻ → Cu(s)

Step 2: Calculate the number of electrons required in the reduction of 0.600 moles of Cu²⁺

We will use the following relationships:

• 1 mole of Cu²⁺ requires 2 moles of electrons

• 1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number)

$0.600molCu^{2+} \times \frac{2mole^{-} }{1molCu^{2+} } \times \frac{6.02 \times 10^{23}e^{-} }{1mole^{-}} = 7.22 \times 10^{23}e^{-}$

Step 3: Calculate the Coulombs corresponding to $7.22 \times 10^{23}e^{-}$

We will use the following relationships:

• 1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number)

• 1 mole of electrons has a charge of 96486 C (Farady's constant)

$7.22 \times 10^{23}e^{-} \times \frac{1 mol e^{-}}{6.02 \times 10^{23}e^{-}} \times \frac{96486C}{1 mol e^{-}} =1.16 \times 10^{5}C$