Question

(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: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\mathrm{e}}=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\text {red }}^{e}=+0.49 \mathrm{~V} \end{array} $$ (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.

Short answer

The discharge and charge reactions for a nickel-cadmium battery are: Discharge: \(Cd(s) + NiO(OH)(s) + H_2O(l) \rightarrow Cd(OH)_2(s) + Ni(OH)_2(s)\) Charge: \(Cd(OH)_2(s) + Ni(OH)_2(s) \rightarrow Cd(s) + NiO(OH)(s) + H_2O(l)\) The standard emf of the cell is calculated as \(E_{cell}^{\circ} = (+0.49\,\text{V}) - (-0.76\,\text{V}) = +1.25\,\text{V}\). The difference between this value and the typical emf value (+1.30 V) is due to non-ideal conditions, impurities, or concentration differences in the actual battery. The equilibrium constant for the overall nicad reaction based on the typical emf value is approximately \(2.03 \times 10^{25}\).

Step-by-step solution

Discharge Reaction

For the discharge reaction, the nickel and cadmium reactants get reduced and oxidized, respectively. Write the reaction as: $$ Cd(s) + NiO(OH)(s) + H_2O(l) \rightarrow Cd(OH)_2(s) + Ni(OH)_2(s) $$

Charge Reaction

For the charge reaction, simply reverse the discharge reaction: $$ Cd(OH)_2(s) + Ni(OH)_2(s) \rightarrow Cd(s) + NiO(OH)(s) + H_2O(l) $$ #b) Calculate the standard emf of the cell using the given reduction potentials#

Calculate the standard emf

To find the standard emf, we have to combine the given half-cell reduction potentials. Take into consideration that for the complete cell, the cadmium half-reaction should be reversed (oxidation). Then, add the potentials: $$ E_{cell}^{\circ} = E_{red}(Ni) - E_{red}(Cd) = (+0.49\,\text{V}) - (-0.76\,\text{V}) = +1.25\,\text{V} $$ #c) Explain the difference between the calculated emf and the typical emf value for a nicad battery#

Difference in emf values

The calculated emf value is +1.25 V, while the typical value is +1.30 V for a nicad battery. This difference can arise due to non-ideal conditions, impurities, or the effect of concentration differences in the actual battery as opposed to the standard conditions used for calculating reduction potentials. #d) Calculate the equilibrium constant for the overall nicad reaction based on the typical emf value#

Calculate the equilibrium constant

Use the Nernst equation to calculate the equilibrium constant (K) of the overall reaction: $$ E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln{K} $$ At equilibrium, \(E_{cell} = 0\), and in this case, \(n = 2\) because there are 2 electrons involved in the reaction. Plug in the typical emf value (+1.30 V), the gas constant R (\(8.314\, J\, mol^{-1}\, K^{-1}\)), Faraday's constant F (\(96485\, C\, mol^{-1}\)), and the temperature (assume standard conditions, T = 298 K): $$ 0 = 1.30\,V - \frac{(8.314\, J\,mol^{-1}\,K^{-1})(298\,K)}{(2)(96485\, C\,mol^{-1})} \ln{K} $$ Solve for the equilibrium constant K: $$ K = e^{\frac{2(1.30\,V)(96485\, C\,mol^{-1})}{(8.314\, J\,mol^{-1} \,K^{-1})(298\,K)}} = 2.03 \times 10^{25} $$ So, the equilibrium constant for the overall nicad reaction based on the typical emf value is approximately \(2.03 \times 10^{25}\).

Key concepts

Reduction Potentials

In electrochemistry, reduction potentials play a crucial role in determining how easily a substance gains electrons and gets reduced. For the nickel-cadmium (nicad) battery, we are given specific reduction potentials for two reactions:

• The reduction of cadmium from cadmium hydroxide with a potential of \(-0.76 \, \text{V}\)
• The reduction of nickel oxyhydroxide to nickel hydroxide with a potential of \(+0.49 \, \text{V}\)

Reduction potentials reflect the tendency of a chemical species to acquire electrons. The more positive the value, the greater the affinity for electrons and the stronger the oxidizing power. In our case, nickel is becoming reduced and cadmium is oxidized, meaning we'll need to reverse the cadmium reaction to convert this information into practical use for calculating the cell's electromotive force.

Standard EMF Calculation

Calculating the standard electromotive force (emf) of a cell is a fundamental task in electrochemical studies. It helps predict the ability of a battery to do work under standard conditions. For our nicad battery, we must employ the standard reduction potentials provided to derive the standard emf.To calculate the emf (\(E^{\circ}_{cell}\)), we need to take the reduction potential of the nickel half-reaction and subtract the reduction potential of cadmium half-reaction (after reversing it, since it acts as an oxidizer):\[E^{\circ}_{cell} = E_{red}(Ni) - E_{red}(Cd)\]Substitute the given values:\[E^{\circ}_{cell} = (+0.49 \, \text{V}) - (-0.76 \, \text{V}) = +1.25 \, \text{V}\]This indicates the cell can ideally produce a voltage of 1.25 V under standard conditions. However, actual operating conditions can differ, leading to discrepancies between calculated and real-life emf values.

Nernst Equation

The Nernst Equation allows us to relate the standard emf of a reaction to its equilibrium constant and real-world conditions. This equation is essential for understanding how factors like ion concentrations affect cell potential. It's defined as:\[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln{Q}\]Where

• \(E_{cell}\) is the cell potential under non-standard conditions,
• \(n\) is the number of electrons,
• \(R\) is the gas constant \(8.314 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\)
• \(T\) is the temperature in Kelvin,
• \(F\) is Faraday's constant \(96485 \ \, \text{C} \cdot \text{mol}^{-1}\)
• \(Q\) is the reaction quotient.

Additionally, at equilibrium, \(E_{cell} = 0\), and the Nernst Equation can solve for the equilibrium constant, \(K\):\[ 0 = E^{\circ}_{cell} - \frac{RT}{nF} \ln{K}\]This relationship is used in our example to find \(K\) using an actual emf of \(1.30 \, \text{V}\). The computed equilibrium constant, \(K = 2.03 \times 10^{25}\), reflects a very favorable reaction under those conditions. The Nernst equation hence bridges the gap between the ideal world of standard conditions and the practical realities of battery operation.