What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{KF}\) solution b. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{MgI}_{2}\) solution

For a \(1.0 M \mathrm{KF}\) solution, we have both \(\mathrm{K}^+\) ions and \(\mathrm{F}^-\) ions. The anode is where oxidation takes place while the cathode is where reduction takes place. At the anode (oxidation), the possible reactions include: -2 \(\mathrm{F}^-\) (aq) → \(\mathrm{F}_{2}\) (g) + 4 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{K}^+\) (aq) + e\(^-\) → \(\mathrm{K}\) (s) - 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq) Using standard reduction potentials, we can determine that fluorine has a higher reduction potential than water, so it will be easier for \(\mathrm{F}^-\) ions to be oxidized. Meanwhile, potassium's reduction potential is lower than that of water, so it will be harder to reduce \(\mathrm{K}^+\) ions compared to the reduction of water. So in the \(1.0 M \mathrm{KF}\) solution, the reactions are: Anode (oxidation): 2 \(\mathrm{F}^-\) (aq) → \(\mathrm{F}_{2}\) (g) + 4 e\(^-\) Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq)

For a \(1.0 M \mathrm{CuCl}_{2}\) solution, we have both \(\mathrm{Cu}^{2+}\) ions and \(\mathrm{Cl}^-\) ions. As before, we must determine the reactions at the anode (oxidation) and cathode (reduction). At the anode (oxidation), the possible reactions include: - 2 \(\mathrm{Cl}^-\) (aq) → \(\mathrm{Cl}_{2}\) (g) + 2 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{Cu}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Cu}\) (s) The most likely reactions to occur are: Anode (oxidation): 2 \(\mathrm{Cl}^-\) (aq) → \(\mathrm{Cl}_{2}\) (g) + 2 e\(^-\) Cathode (reduction): \(\mathrm{Cu}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Cu}\) (s)

For a \(1.0 M \mathrm{MgI}_{2}\) solution, we have both \(\mathrm{Mg}^{2+}\) ions and \(\mathrm{I}^-\) ions. Let's determine the reactions at the anode (oxidation) and the cathode (reduction). At the anode (oxidation), the possible reactions include: - 2 \(\mathrm{I}^-\) (aq) → \(\mathrm{I}_{2}\) (s) + 2 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{Mg}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Mg}\) (s) - 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq) Using standard reduction potentials, we can determine that magnesium's reduction potential is lower than that of water, so it will be harder to reduce \(\mathrm{Mg}^{2+}\) ions compared to the reduction of water. So in the \(1.0 M \mathrm{MgI}_{2}\) solution, the reactions are: Anode (oxidation): 2 \(\mathrm{I}^-\) (aq) → \(\mathrm{I}_{2}\) (s) + 2 e\(^-\) Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq)