Question

03/viewContent/39478147/View Unit Activity 8: Oxidation-Reduction in Batteries In Chapter 8 we learn about oxidation-reduction reactions. This type of reaction is very prevalent in our everyday life, usually in the form of batteries. There are currently many different types of batteries. Their sizes and the materials they are made of vary depending on what they are being used for and how much energy is being supplied. In this activity, you will contrast and compare two different types of batteries that are commonly used. You will discuss them in terms of the chemical reaction that is used and also the amount of energy they provide. 1 Choose two batteries from the following: a. Nickel-metal hydride cell b. Ni-Cad cell c. Lithium-iodine cell d. Silver oxide cell e. Lithium-Fes, battery f. Lithium-SO, cell 2. Choose 2 of the batteries from the list. How much energy per charge (or voltage) does each battery you chose have? 3. What are some typical uses for each battery? 4. For 1 of the batteries, give the balanced oxidation and reduction half-reactions. (Use both your textbook and online resources to determine more information about each type of battery) In the half-teactions in 4. determine which element is being oxidized and which element is being reduced

Answer 1

(i) Ni-Cad cell: Here Cadmium acts as an anode, whereas Nickel acts as a cathode.

Reaction at anode: Cs(s) ----> Cd²⁺(aq) + 2e^-; E^o_anode = -0.403 V

Reaction at cathode: Ni²⁺(aq) + 2e^- ----> Ni(s); E^o_cathode = -0.257 V

Now, E^o_cell = E^o_cathode - E^o_anode = -0.257 - (-0.403) = 0.146 V

Note: Here, both the anode and cathode are the metal-electrodes.

(ii) Lithium-iodine cell: Here Lithium acts as an anode, whereas Iodine acts as a cathode.

Reaction at anode: Li(s) ----> Li⁺(aq) + e^-; E^o_anode = -3.0401 V

Reaction at cathode: I₂(s) + 2e- ----> 2I^-(aq); E^o_cathode = 0.5355 V

Now, E^o_cell = E^o_cathode - E^o_anode = 0.5355 - (-3.0401) = 3.5756 V

Note: Here, anode is metal-electrode, whereas cathode is nonmetal-electrode.