Q21.149CP - Chemistry: Molecular Nature Of Matter And Change

Question

The zinc-air battery is a less expensive alternative for silver batteries in hearing aids. The cell reaction is

$\mathbf{2}\mathbf{Zn}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }\mathbf{2}\mathbf{ZnO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

A new battery weighs  $\mathbf{0}\mathbf{.}\mathbf{275}$ g. The zinc accounts for exactly  $\frac{\mathbf{1}}{\mathbf{10}}$ of the mass, and the oxygen does not contribute to the mass because it is supplied by the air.

(a) How much electricity (in C) can the battery deliver?

(b) How much free energy (in J) is released if ${\mathbf{E}}_{\mathbf{cell}}$  is  $\mathbf{1}\mathbf{.}\mathbf{3}$ V?

Step-by-Step Solution

Answer

(a) The amount of electricity that a battery can deliver in terms of C is obtained as: $81.18\mathrm{C}$.

(b) The amount of free energy released in terms of J is obtained as: $-2.51\times {10}^5\mathrm{J}$.

Step 1: Chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate a new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Step 2: Subpart (a)

As the mass of Zn is $\frac{1}{10}$of $0.275$, , the charge of the reaction may be calculated by dividing the mass by its molar mass, translating the moles of Zn to moles of electron, and multiplying by the Faraday constant. In the oxidation of $\mathrm{Zn}\to {\mathrm{Zn}}^{2+}$, two electrons are involved.

$0.0275\mathrm{gZn}\times \frac{1\mathrm{molZn}}{65.38\mathrm{gZn}}\frac{2{\mathrm{mole}}^{-}}{1\mathrm{molZn}}\times \frac{96,500\mathrm{C}}{1{\mathrm{mole}}^{-}}=81.18\mathrm{C}$

Therefore, the electricity battery delivers is $81.18\mathrm{C}$.

Step 3: Subpart (b)

The equation $\mathrm{\Delta G}=-{\mathrm{nFE}}_{\mathrm{cell}}$ may be used to compute the $\mathrm{\Delta G}$.

$\begin{array}{c}\mathrm{\Delta G}=-{\mathrm{nFE}}_{\mathrm{cell}}\\ =-2\times 96,500\times 1.3\\ =-2.51\times {10}^5\mathrm{J}\end{array}$

Therefore, the amount of free energy released is $-2.51\times {10}^5\mathrm{J}$ .