First, we need to find the total charge transferred during the 15 minutes. We can do this using the formula:Charge Q = Current I Time t Given that the current I is 2.0 A and the time t is 15 minutes, we need to convert the time to seconds:15 minutes 60 seconds/minute = 900 secondsNow we can calculate the charge:Q = 2.0 A 900 s = 1800 CNext, we need to find the number of moles of electrons transferred. We can do this using the Faraday constant F , which is 96,485 C/mol:Moles of electrons n = Charge Q / Faraday constant F n = 1800 C / 96,485 C/mol 0.01865 molNow, we know that in the redox reaction of CuSO4, copper Cu is reduced from Cu to Cu, which means that each Cu ion gains 2 electrons. Therefore, we need to find the number of moles of Cu ions reduced:Moles of Cu ions = Moles of electrons / 2Moles of Cu ions 0.01865 mol / 2 0.009325 molFinally, we can find the total number of electrons transferred by multiplying the moles of electrons by Avogadro's number 6.022 10 particles/mol :Total number of electrons 0.01865 mol 6.022 10 electrons/mol 1.12 10 electronsSo, the total number of electrons transferred when a current of 2.0 A flows through a solution of CuSO4 for 15 minutes is approximately 1.12 10 electrons.