Answer :
The concentration of Cu²⁺ in the solution after 15.27 hours of electrolysis under a current of 1.55 A is 2.79 M.
The equation for the electrolysis of Cu²⁺ is: Cu²⁺(aq) + 2e⁻ → Cu(s)
From this equation, we can see that for every 2 electrons transferred, 1 mole of Cu²⁺ is reduced to form 1 mole of Cu. The amount of charge transferred during electrolysis can be calculated using the formula: Q = It
Since 2 electrons are required to reduce 1 mole of Cu²⁺, the number of moles of Cu²⁺ reduced can be calculated as: moles of Cu²⁺ = (83,565 C) / (2 × 96,485 C/mol)
moles of Cu²⁺ = 0.433 mol
The new volume of the solution is still 1.00 L. Therefore, the new concentration of Cu²⁺ can be calculated as: new [Cu²⁺] = (0.433 mol) / (1.00 L). New [Cu²⁺] = 0.433 M
The concentration of Cu²⁺ in a 1.00 L solution with a concentration of 6.28 M was found to be 0.433 M after 15.27 hours of electrolysis under a current of 1.55 A. This was calculated by first finding the amount of charge transferred using Q = It, and then using the equation for the electrolysis of Cu²⁺ to calculate the number of moles of Cu²⁺ reduced.
Assuming a negligible change in volume, the concentration of Cu²⁺ after electrolysis was found to be approximately 2.79 M.
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