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(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metalhydride batteries over the nickel-cadmium batteries?

Short Answer

Expert verified
A battery made with cadmium metal instead of zinc would have a higher cell emf, as the standard reduction potential of cadmium is higher by 0.36 V compared to zinc. Nickel-metal hydride (Ni-MH) batteries provide an environmental advantage over nickel-cadmium (Ni-Cd) batteries, as they do not contain the toxic heavy metal cadmium, reducing the risk of heavy metal toxicity and making it easier to recycle, thus providing a more environmentally friendly option with fewer human health risks.

Step by step solution

01

Analyze the standard reduction potentials of cadmium and zinc

To understand the effect of using cadmium instead of zinc in an alkaline battery, we need to know the standard reduction potential of both elements. The standard reduction potential tells us the tendency of an element to be reduced (gain electrons) or oxidized (lose electrons). A higher reduction potential means the element is more likely to gain electrons and act as an oxidizing agent. The standard reduction potentials of zinc and cadmium are as follows: Zn: \(Zn^{2+} + 2e^- \rightarrow Zn\), \(E^0 = -0.76 V\) Cd: \(Cd^{2+} + 2e^- \rightarrow Cd\), \(E^0 = -0.40 V\) It's essential to know the difference between their reduction potentials as that would determine the change in cell emf.
02

Calculate the difference in standard reduction potentials

Now, let's compute the difference in standard reduction potentials between these two elements to see how the cell emf is affected. Difference: \((E^0_{Cd} - E^0_{Zn}) = (-0.40 V) - (-0.76 V) = 0.36 V\) The standard reduction potential of cadmium is higher by 0.36 V compared to zinc. This indicates that a battery made with cadmium would have a higher cell emf than one made with zinc.
03

Discuss the environmental advantage of nickel-metal hydride batteries over nickel-cadmium batteries

Nickel-cadmium (Ni-Cd) batteries contain the heavy metal cadmium, which is highly toxic when ingested or inhaled. When Ni-Cd batteries are not properly disposed of, they can leak cadmium into the environment, causing soil, water, and air pollution. This could have severe health consequences for humans and wildlife. On the other hand, nickel-metal hydride (Ni-MH) batteries do not contain cadmium, thus reducing the risk of heavy metal toxicity. The absence of cadmium also makes Ni-MH batteries easier to recycle compared to Ni-Cd batteries. Nickel-metal hydride batteries, therefore, provide a more environmentally friendly alternative, with fewer human health risks and a smaller ecological footprint.

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Most popular questions from this chapter

Derive an equation that directly relates the standard emf of a redox reaction to its equilibrium constant.

This oxidation-reduction reaction in acidic solution is spontaneous: \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow\) \(5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

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