Chapter 17: Q9P (page 428)

Fundamentals of Electrolysis
17-9. A lead-acid battery in a car has six cells in series, each delivering close to2.0Vfor a total of 12V when the battery is discharging. Recharging requires $~ 2.4 V$ per cell, or $~ 14 V$ for the entire battery. ⁵⁵Explain these observations in terms of Equation 17-6.

Expert verified

Electrical losses decrease the magnitude of the voltage in the cell and increase the magnitude of the voltage in the opposite direction.

Step by step solution

Electrogravimetry is a technique for separating and quantifying ions of material, most often a metal.
The analyte solution is electrolyzed during this procedure.
The analyte is deposited on the cathode as a result of the electrochemical reduction.

The equation $17 - 6$ is:
$E = E (cathode) - E (anode) - I . R - overpotentials$
When current does not flow, the potential is E (cathode) - E (anode) = 2.2 V .
When there is current, the ohmic loss and overpotentials decrease the output of the cell 2.0V , and a net cell voltage is then 2.0V.
The cell can be recharged at a very low current when the voltage $2.2 V$ is used in the opposite direction.
If we want to charge at a significant rate, we need additional voltage to overcome ohmic loss and overpotentials.
The recharge is then required $- 2.4 V$ .
Electrical losses decrease the magnitude of the voltage in the cell and increase the magnitude of the voltage in the opposite direction.