Answer:
[tex]7.22 \times 10^{23}e^{-}[/tex]
[tex]1.16 \times 10^{5}C[/tex]
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:
[tex]0.600molCu^{2+} \times \frac{2mole^{-} }{1molCu^{2+} } \times \frac{6.02 \times 10^{23}e^{-} }{1mole^{-}} = 7.22 \times 10^{23}e^{-}[/tex]
Step 3: Calculate the Coulombs corresponding to [tex]7.22 \times 10^{23}e^{-}[/tex]
We will use the following relationships:
[tex]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[/tex]
