What mass of Cu would be produced by the cathodic reduction of \(\mathrm{Cu}^{2+}\) when 1.60A of current passes through a solution of \(\mathrm{CuSO}_{4}\)...

The reduction reaction that occurs at the cathode during the electrolysis of CuSO4" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-1-Frame">4, is: Cu2+" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-2-Frame">2+ + 2e- -> Cu(s) From this, we can see that each Cu2+ ion requires two electrons to be reduced to copper metal. Given the current (I = 1.60 A), time (t = 1 hour = 3600 s), and Faraday's constant (F = 96500 C/mol), we can calculate the total amount of charge that passes through the solution: Q = I*t = 1.60 A * 3600 s = 5760 C Using Faraday's law, we can relate the amount of charge that passes through the solution to the number of moles of electrons transferred during the reduction reaction: n = Q/F = 5760 C / 96500 C/mol = 0.0597 mol e- Since each Cu2+ ion requires 2 electrons to be reduced to copper metal, the number of moles of copper produced is half the number of moles of electrons transferred: mol Cu = 0.0597 mol e- / 2 = 0.0299 mol Cu Finally, we can convert the moles of copper produced to grams using the molar mass of copper: mass Cu = 0.0299 mol Cu * 64 g/mol = 1.91 g Therefore, the answer is 1.91 g of Cu produced. is correct.