CHEMISTRY HELP PLEASE: electrochemistry?

Question

A) A "dead" 12-Volt lead/acid storage battery has 4.40 g of PbSO4 (molar mass = 303.3 g), deposited on its electrodes. To recharge it, a ``trickle charger" is attached to the Pb anode that supplies 0.280 amperes of current at a voltage of 13.0 V. Calculate how much time it would take to convert all the PbSO4 back into SO42- ions in solution.

B) How long would it take to electrodeposit all the Cu2+ in 0.300 L of 0.250 M CuSO4 solution with an applied potential difference of 0.275 V and a current of 2.65 amperes? Enter your answer in minutes.

thanks a lot!

Answer 1

A)
4.40 g of PbSO4 (molar mass = 303.3 g) = 0.01451 mole of PbSO4
==> 0.02902 mole of e- needed to convert all SO4(2-)
Since the charge of each electron is: -1.602x10^-19 C, the magnitude of the charge of 0.02902 mole of e- is:
0.02902*6.022x10^23*1.602x10^-19 = 2800 (C)
Remember 1 Ampere = 1 C/s. So required time is;
2800C / (0.280C/s) = 10000s = 166.7 min = 2.778 hours.
(By the way, 6.022x10^23*1.602x10^-19 = 96485(C) is One Faraday )

B) Using the same way:
0.300*0.250*2*96485/2.65 = 5461(s) = 91.0 min

Answer 2

for part one all you have to do is find electron moles which is moles of pbso4*2 becase in equation 2e- are transfered.
then simple plug in the values to equation: Molese- = Current * Time / Farrday constant
so Time= 96485*0.02902 /0.280