Suppose 650.mmol of electrons must be transported from one side of an electrochemical cell to another in 10.0 minutes. Calculate the size of electric current that must flow.

Answer:

Suppose 650.mmol of electrons must be transported from one side of an electrochemical cell to another in 10.0 minutes. Calculate the size of electric current that must flow.

Current = Charge / time

Charge = mol of e- * charge of electron = (0.650 mol of e- ) * (96500 C/mol of e- ) = 62725 C

time = 10 min = 10 min * 60 s / min = 600 s

I = C/t = (62725)/600

I = 104.54 C/s

recall that 1 C/s = 1 Amp

I = 104.54 A