(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{aligned} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

The discharge reaction involves the conversion of chemical energy stored in the battery to electrical energy. For a nickel-cadmium battery, the discharge reaction can be written as: $$ \mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s) $$

The charge reaction involves recharging the battery and converting electrical energy back into chemical energy. For a nickel-cadmium battery, the charge reaction is the reverse of the discharge reaction: $$ \mathrm{Cd}(\mathrm{OH})_{2}(s)+2\mathrm{Ni}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cd}(s)+2\mathrm{NiO}(\mathrm{OH})(s)+2\mathrm{H}_{2}\mathrm{O}(l) $$ #b) Standard Emf of the Cell#

We can calculate the standard emf of the cell using the given reduction potentials. Remember that the overall cell potential (\(E_\text{cell}^\circ\)) is the difference between the reduction potential of the cathode (\(E_\text{red,cathode}^\circ\)) and the reduction potential of the anode (\(E_\text{red,anode}^\circ\)): $$ E_\text{cell}^\circ = E_\text{red,cathode}^\circ - E_\text{red,anode}^\circ $$ Using the given reduction potentials: $$ E_\text{cell}^\circ = (+0.49 \,\text{V}) - (-0.76 \,\text{V}) = +1.25 \,\text{V} $$ #c) Difference in Experimental and Calculated Emf#

The experimental emf value given for a typical nicad battery is +1.30 V, which is different from the calculated emf value of +1.25 V. This difference can be attributed to real-world factors such as the specific compositions of the battery components, temperature, and impurities that can affect the reduction potentials and overall cell potential. These factors are usually not accounted for in the theoretical calculation of standard emf. #d) Equilibrium Constant#

We can calculate the equilibrium constant (\(K_\text{eq}\)) for the overall nicad reaction based on the given typical emf value, using the Nernst equation: $$ E_\text{cell} = E_\text{cell}^\circ - \frac{RT}{nF} \ln K_\text{eq} $$ In this exercise, we are given the experimental value of the cell potential (\(E_\text{cell} = 1.30 \,\text{V}\)) and we know the number of moles of electrons transferred (\(n=2\)) in the reaction. We can rearrange the Nernst equation to solve for \(K_\text{eq}\): $$ K_\text{eq} = e^{\frac{nFE_\text{cell}}{RT}} $$ Using \(E_\text{cell} = 1.30 \,\text{V}\), \(n=2\), and standard values for the gas constant (\(R = 8.314 \,\text{J/mol}\,\text{K}\)) and Faraday's constant (\(F =96485 \,\text{C/mol}\)) at room temperature (298 K): $$ K_\text{eq} = e^{\frac{(2)(96485\,\text{C/mol})(1.30\,\text{V})}{(8.314\,\text{J/mol}\,\text{K})(298\,\text{K})}} $$ $$ K_\text{eq} \approx 1.5 \times 10^{18} $$ So the equilibrium constant for the overall nicad reaction is approximately \(1.5 \times 10^{18}\).