What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M\) NiBr \(_{2}\) solution b. \(1.0 M\) AlF solution c. \(1.0 M\) MnI\(_{2}\) solution

Electrolysis is a process that uses an electric current to drive a non-spontaneous redox reaction. In electrolysis, the anode is the positive electrode, where oxidation takes place, and the cathode is the negative electrode, where reduction takes place.

For each of the given electrolyte solutions, we will need to look up their reduction potentials from a standard reduction potentials table. These values will help us identify which species are most likely to be reduced or oxidized at the electrodes. For the cases in question: a. NiBr\(_2\): Ni\(^{2+}\) and Br\(^{-}\) will be the ions in solution. -Ni\(^{2+}\) reduction potential: Ni\(^{2+}\) + 2e\(^{-}\) → Ni ; E° = -0.25V -Br\(^{-}\) reduction potential: 2Br\(^{-}\) → Br\(_2\) + 2e\(^{-}\) ; E° = +1.09V b. AlF: Al\(^{3+}\) and F\(^{-}\) will be the ions in solution. -Al\(^{3+}\) reduction potential: Al\(^{3+}\) + 3e\(^{-}\) → Al ; E° = -1.66V -F\(^{-}\) reduction potential: 2F\(^{-}\) → F\(_2\) + 2e\(^{-}\) ; E° = +2.87V c. MnI\(_2\): Mn\(^{2+}\) and I\(^{-}\) will be the ions in solution. -Mn\(^{2+}\) reduction potential: Mn\(^{2+}\) + 2e\(^{-}\) → Mn ; E° = -1.18V -I\(^{-}\) reduction potential: 2I\(^{-}\) → I\(_2\) + 2e\(^{-}\) ; E° = +0.54V

To determine the reactions occurring at the electrodes, we will choose the species with the highest reduction potential to be reduced at the cathode and the species with the lowest reduction potential to be oxidized at the anode. For each solution, do the following: a. NiBr\(_2\) solution -Cathode reaction (reduction): Ni\(^{2+}\) + 2e\(^{-}\) → Ni ; E° = -0.25V -Anode reaction (oxidation): 2Br\(^{-}\) → Br\(_2\) + 2e\(^{-}\) ; E° = +1.09V b. AlF solution -Cathode reaction (reduction): Al\(^{3+}\) + 3e\(^{-}\) → Al ; E° = -1.66V -Anode reaction (oxidation): 2F\(^{-}\) → F\(_2\) + 2e\(^{-}\) ; E° = +2.87V c. MnI\(_2\) solution -Cathode reaction (reduction): Mn\(^{2+}\) + 2e\(^{-}\) → Mn ; E° = -1.18V -Anode reaction (oxidation): 2I\(^{-}\) → I\(_2\) + 2e\(^{-}\) ; E° = +0.54V So the reactions occurring at the electrodes during the electrolysis of each solution are: a. NiBr\(_2\) solution -Cathode: Ni\(^{2+}\) + 2e\(^{-}\) → Ni -Anode: 2Br\(^{-}\) → Br\(_2\) + 2e\(^{-}\) b. AlF solution -Cathode: Al\(^{3+}\) + 3e\(^{-}\) → Al -Anode: 2F\(^{-}\) → F\(_2\) + 2e\(^{-}\) c. MnI\(_2\) solution -Cathode: Mn\(^{2+}\) + 2e\(^{-}\) → Mn -Anode: 2I\(^{-}\) → I\(_2\) + 2e\(^{-}\)