Even though the toxicity of cadmium has become a concern, nickel-cadmium (nicad) batteries are still used commonly in many devices. The overall cell reaction is:

$\mathbf{Cd}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{NiO}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{Ni}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{Cd}\mathbf{(}\mathbf{OH}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

A certain nicad battery weighs$\mathbf{18}\mathbf{.}\mathbf{3}$g and has a capacity of $\mathbf{300}$mAh (that is, the cell can store charge equivalent to a current of$\mathbf{300}$mA flowing for 1 h). (a) What is the capacity of this cell in coulombs? (b) What mass of reactants is needed to deliver $\mathbf{300}$mAh? (c) What percentage of the cell mass consists of reactants?

To calculate the cell's capacity, multiply the duration by the current in mA.

The half responses are as follows:

$\begin{array}{c}\mathrm{Oxidation}:\mathrm{Cd}\to {\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-}\\ \mathrm{Reduction}:2\mathrm{Ni}\left(\mathrm{OH}\right)+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to \mathrm{Ni}\left(\mathrm{OH}\right)+2\mathrm{O}{\mathrm{H}}^{-}\end{array}$

Two electrons are involved in each half-reaction. To calculate the moles of electrons, divide the charge by Faraday's constant to get the mass of the reactants.

$1,080\mathrm{C}\times \frac{{\mathrm{mole}}^{-}}{96,500\mathrm{C}}=0.0112{\mathrm{mole}}^{-}$

To find the mass of each reagent, convert the moles of electron to moles of reagent and multiply by the molar masses.

The mass Cd is:

role="math" localid="1663933619869" $\begin{array}{c}\mathrm{gm}\mathrm{of}\mathrm{Cd}=0.0112{\mathrm{mole}}^{-}\times \frac{1\mathrm{molCd}}{2{\mathrm{mole}}^{-}}\times \frac{112.4\mathrm{g}}{1\mathrm{molCd}}\\ =0.629\end{array}$

The mass NiO(OH) is:

$\begin{array}{c}\mathrm{gm}\mathrm{of}\mathrm{Ni}\left(\mathrm{OH}\right)=0.0112{\mathrm{mole}}^{-}\times \frac{1\mathrm{molNiO}\left(\mathrm{OH}\right)}{1{\mathrm{mole}}^{-}}\times \frac{91.70\mathrm{g}}{1\mathrm{molNiO}\left(\mathrm{OH}\right)}\\ =1.028\end{array}$

The mass is:

$\begin{array}{c}{\mathrm{gmofH}}_2\mathrm{O}=0.0112{\mathrm{mol}}^{-}\times \frac{1{\mathrm{molH}}_2\mathrm{O}}{1{\mathrm{mole}}^{-}}\times \frac{18.01\mathrm{g}}{1{\mathrm{molH}}_2\mathrm{O}}\\ =0.202\mathrm{g}\end{array}$

$\begin{array}{c}{\mathrm{mass}}_{\mathrm{total}}={\mathrm{mass}}_{\mathrm{Cd}}+{\mathrm{mass}}_{{\mathrm{H}}_2\mathrm{O}}+{\mathrm{mass}}_{\mathrm{NiO}\left(\mathrm{OH}\right)}\\ =0.629+1.028+0.202\\ =1.859\mathrm{g}\end{array}$

Therefore, the mass is:

$\begin{array}{c}\mathrm{Cd}=0.629\mathrm{g}\\ \mathrm{Ni}\left(\mathrm{OH}\right)=1.028\mathrm{g}\\ {\mathrm{H}}_2\mathrm{O}=0.202\mathrm{g}\\ \mathrm{Total}=1.859\mathrm{g}\end{array}$

Multiply the mass ratio of the reactants over the total mass by to obtain the ratio.

$\frac{1.859}{18.3}\times 100=10.2\mathrm{percent}$

Therefore, the cell mass contains:$10.2\%$.