Consider two electrolytic cells A and B. Cell A contains a \(0.20 \mathrm{M} \mathrm{CoSO}_{4}\) solution and platinum electrodes. Cell B differs from cell A only in that cobalt metals are used as electrodes. In each case, a current of \(0.20 \mathrm{~A}\) is passed through the cell for \(1.0 \mathrm{~h}\). (a) Write equations for the half-cell and overall cell reactions for these cells. (b) Calculate the products formed (in grams) at the anode and cathode in each case.

Understanding electrode half-cell reactions is crucial when studying electrolytic cells. These reactions occur at the electrodes, where either reduction (gain of electrons) or oxidation (loss of electrons) takes place. In the electrolytic cells A and B discussed, the key difference lies in the type of electrodes used.

In Cell A, with inert platinum electrodes, the cobalt ions in the solution undergo reduction at the cathode, producing solid cobalt metal. The reaction is as follows: \[ Co^{2+}(aq) + 2e^- \rightarrow Co(s) \]. Since platinum electrodes do not participate in the reaction, this is the only half-cell reaction taking place.

In contrast, Cell B uses cobalt metal electrodes. Here, we have two half-cell reactions. At the anode, the cobalt metal is oxidized to cobalt ions: \[ Co(s) \rightarrow Co^{2+}(aq) + 2e^- \]. Simultaneously, at the cathode, cobalt ions are reduced back to cobalt metal. This demonstrates the dynamic nature of electrode half-cell reactions where different electrodes can dramatically change the processes occurring within an electrolytic cell.

Electrolysis calculations help quantify the chemical changes occurring during electrolysis. These calculations are based on the laws of electrolysis, which include Faraday's laws of electrolysis relating charge, moles of electrons, and the amount of substance electrolyzed.

To calculate the mass of substance formed or consumed at an electrode, you would use the following relationships: \[ \text{mass} = \frac{\text{charge} \times \text{molar mass}}{n \times F} \], where 'charge' is the current (in amperes) multiplied by time (in seconds), 'n' is the number of moles of electrons, and 'F' is Faraday's constant (96485 C/mol).

In our exercise, with a current of 0.20 A for 1 hour (which equals 3600 seconds), we can calculate the amount of cobalt deposited or dissolved at the electrodes. This kind of calculation allows for precise control over the electrolytic processes, enabling the determination of the mass of products at the anode and cathode.

Oxidation and reduction are chemical processes that are central to understanding electrolytic cell reactions. Oxidation involves the loss of electrons, while reduction is the gain of electrons. These two processes always occur together since the electrons transferred during oxidation are gained by another species during reduction.

Oxidation occurs at the anode, and in our example with Cell B, where cobalt electrodes are used, the cobalt metal is oxidized. The oxidation half-reaction is given by: \[ Co(s) \rightarrow Co^{2+}(aq) + 2e^- \]. Reduction takes place at the cathode and in both Cell A and Cell B, cobalt ions in the solution are reduced to solid cobalt at the cathode: \[ Co^{2+}(aq) + 2e^- \rightarrow Co(s) \].

These two processes illustrate the movement and transfer of electrons which is a fundamental concept in electrochemistry and drives the operation of electrolytic cells.