Question

During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to$\text{ }1.139\text{ g/mL }$. Sulphuric acid of density $\text{ }1.294\text{ g/mL }$ is $39{\scriptstyle{}^{0}/{}_{0}}$ by weight and that of density is $20{\scriptstyle{}^{0}/{}_{0}}$ by weight. The battery holds $\text{ 3}\text{.5 L}$ of the acid and the volume remains practically constant during the discharge. Find the number of ampere-hours for which the battery must have been used.

Answer 1

Faraday's law has given the relationship between charge passed through the electrolyte and the amount of substance deposited on the electrode. The amount of mass liberated or deposited at the electrode is directly proportional to the amount of charge passed through the solution and chemical equivalent weight.

Complete step by step answer:
Let’s first write the data proved to us.
The initial density of sulfuric acid is
$\begin{aligned} & \text{ }{{\text{d}}_{\text{1}}}\text{=1}\text{.294g/mL } \\\ & \text{mass }{\scriptstyle{}^{0}/{}_{0\text{ }}}=\text{ }39{\scriptstyle{}^{0}/{}_{0}} \\\ \end{aligned}$
The final density of sulfuric acid is
$\begin{aligned} & \text{ }{{\text{d}}_{2}}\text{=1}\text{.139g/mL } \\\ & \text{mass }{\scriptstyle{}^{0}/{}_{0}}\text{ = }20{\scriptstyle{}^{0}/{}_{0}} \\\ \end{aligned}$
The volume of acid in the battery is 3.5L
We have to find the number of ampere-hours used by the lead storage battery.

Let’s first write down the charging and discharging reactions for the lead storage reaction,
The charging and discharging reaction for the lead storage battery.

& \text{ Pb+SO}_{\text{4}}^{\text{2-}}\to \text{PbS}{{\text{O}}_{\text{4}}}\text{+2}{{\text{e}}^{\text{-}}}\text{ (Charging)} \\\ & \text{ Pb}{{\text{O}}_{\text{2}}}\text{+4}{{\text{H}}^{\text{+}}}\text{+SO}_{\text{4}}^{\text{2-}}\text{+2}{{\text{e}}^{\text{-}}}\to \text{PbS}{{\text{O}}_{\text{4}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O (Discharging) } \\\ \end{aligned}$$ Therefore, the net reaction is , $\text{Pb+Pb}{{\text{O}}_{\text{2}}}\text{+4}{{\text{H}}^{\text{+}}}\text{+2SO}_{\text{4}}^{\text{2-}}\to \text{2PbS}{{\text{O}}_{\text{4}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O}$ The normality of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ and molarity of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is equal. Since, $\text{ 2SO}_{\text{4}}^{\text{2-}}$ requires the$\text{ 2}{{\text{e}}^{\text{-}}}$. Now let's find out the molarity of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ before and after the electrolysis. We are familiar with the formula which relates the molarity of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$with density, mass%, and molecular weight. $$\begin{aligned} & {{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{= mass }{\scriptstyle{}^{0}/{}_{0}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{density }\\!\\!\times\\!\\!\text{ 1000}}{\text{molecular weight}} \\\ & \\\ & \text{-Before electrolysis, } \\\ & \\\ & {{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{=}\dfrac{\text{39}}{\text{100}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{1}\text{.294 }\\!\\!\times\\!\\!\text{ 1000}}{\text{98}} \\\ & \text{ }{{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{=}\dfrac{\text{50466}}{\text{9800}}\text{=5}\text{.15} \\\ \end{aligned}$$ -Similarly, the molarity of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ after hydrolysis, $\begin{aligned} & \text{ }{{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{=}\dfrac{\text{20}}{\text{100}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{1}\text{.139 }\\!\\!\times\\!\\!\text{ 1000}}{\text{98}} \\\ & \text{ }{{\text{M}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}\text{=}\dfrac{\text{22780}}{\text{9800}}\text{=2}\text{.235 } \\\ \end{aligned}$ Now, let's find out the no of moles the before and after molarity of${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ $\begin{aligned} & \text{mole of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=molarity }\\!\\!\times\\!\\!\text{ volume in d}{{\text{m}}^{\text{3}}} \\\ & \\\ & \text{-Before electrolysis, } \\\ & \text{ mole of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=5}\text{.15 }\\!\\!\times\\!\\!\text{ 3}\text{.5=18}\text{.025} \\\ & \\\ & \text{-After electrolysis, } \\\ & \text{mole of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=2}\text{.325 }\\!\\!\times\\!\\!\text{ 3}\text{.5=8}\text{.137 } \\\ \end{aligned}$ Therefore, a mole equivalent i.e. $\text{ }\dfrac{\text{w}}{\text{E}}\text{ of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ used = }18.025-8.137=9.888\text{ }$ We want to find the current passing through the battery. We will use the concept of faraday's law to get the current flowing through the battery. We are passing one faraday of charge through the electrolyte. It means if we are passing the one-mole electron by carrying 1 Faraday of charge and 1gm of equivalent weight is either deposited or liberated on the electrode. In short, by combining the faraday's law we get the relationship between charge passed through the electrolyte, mass deposited or liberated at the electrode, and chemical equivalent weight. The relation is, $\text{ w=}\dfrac{\text{q}}{\text{96500}}\text{ }\\!\\!\times\\!\\!\text{ E }$ $$\begin{aligned} & \dfrac{\text{w}}{\text{E}}\text{=}\dfrac{\text{i }\\!\\!\times\\!\\!\text{ t}}{\text{96500}} \\\ & \\\ & \text{we know charge is given as,} \\\ & \text{ q = i }\\!\\!\times\\!\\!\text{ t } \\\ \end{aligned}$$ (1) Faraday's constant is $\text{ 96500 C mo}{{\text{l}}^{\text{-1}}}$ Let's substitute the values in equation (1). $$\begin{aligned} & \text{i }\\!\\!\times\\!\\!\text{ t=9}\text{.8875 }\\!\\!\times\\!\\!\text{ 96500 since,}\dfrac{\text{w}}{\text{E}}=9.8875 \\\ & \text{i }\\!\\!\times\\!\\!\text{ t=954143}\text{.75 ampere sec} \\\ \end{aligned}$$ We have to find the current flowing through the battery in an hour. Thus convert the unit second to the hour dividing it by 3600. We get, $$\begin{aligned} & \text{i }\\!\\!\times\\!\\!\text{ t=954143}\text{.75 ampere sec} \\\ & \text{i }\\!\\!\times\\!\\!\text{ t=}\dfrac{\text{954143}\text{.75}}{60\times 60}\text{ ampere hour} \\\ & \text{i }\\!\\!\times\\!\\!\text{ t=265}\text{.03 ampere hour} \\\ \end{aligned}$$ Therefore, the number of ampere-hours for which the battery must have been used is a 265.03 ampere-hour. **Note:** The faraday's constant is usually taken as $\text{ }9.65\times {{10}^{4}}\text{ C mo}{{\text{l}}^{\text{-1}}}\text{ }$(coulombs per mole). But always remember another value which is $\text{ 96500 C mo}{{\text{l}}^{\text{-1}}}$.