What is produced at each electrode in the electrolysis of NaBr(aq) and AgF(aq)?
Well, in the electrolysis of NaBr(aq), at the positive electrode (anode), bromine (Br) is produced. And at the negative electrode (cathode), sodium (Na) is produced. So, you can say it's a shocking bromance taking place.
As for the electrolysis of AgF(aq), at the positive electrode (anode), oxygen (O2) is produced. And at the negative electrode (cathode), silver (Ag) is produced. So, you can say it's a breath of fresh air combined with a shiny surprise!
Keep in mind that these reactions are based on aq (aqueous) solutions, so secretion of substances other than the ones I mentioned could also occur, depending on the conditions.
In the electrolysis of NaBr(aq), sodium bromide, the following products are produced at each electrode:
At the anode (positive electrode): Water is oxidized, producing oxygen gas (O2) and releasing electrons. The overall reaction is 2H2O(l) -> O2(g) + 4H+(aq) + 4e-.
At the cathode (negative electrode): Bromide ions (Br-) are reduced, resulting in the formation of bromine (Br2) and gaining electrons. The overall reaction is 2Br-(aq) + 2e- -> Br2(l).
In the electrolysis of AgF(aq), silver fluoride, the following products are produced at each electrode:
At the anode (positive electrode): Water is oxidized, producing oxygen gas (O2) and releasing electrons. The overall reaction is 2H2O(l) -> O2(g) + 4H+(aq) + 4e-.
At the cathode (negative electrode): Fluoride ions (F-) are reduced, resulting in the formation of hydrogen gas (H2) and gaining electrons. The overall reaction is 2F-(aq) + 2e- -> F2(g).
In order to determine what is produced at each electrode during the electrolysis of NaBr(aq) and AgF(aq), we need to understand the process of electrolysis and determine the potential of each half-reaction.
During electrolysis, a direct electric current is passed through an electrolyte to induce a chemical reaction. The electrolyte is a solution that conducts electricity due to the presence of ions. In this case, NaBr(aq) and AgF(aq) are the electrolytes.
Let's break down the electrolysis process into two half-reactions: the oxidation half-reaction (occurs at the anode, the positive electrode) and the reduction half-reaction (occurs at the cathode, the negative electrode). We can then determine the products produced at each electrode based on their standard electrode potentials.
For NaBr(aq):
- Oxidation half-reaction: 2Br^- (aq) -> Br2 (g) + 2e^-
- Reduction half-reaction: 2H2O (l) + 2e^- -> H2 (g) + 2OH^- (aq)
The half-reaction with the higher reduction potential will occur at the cathode, while the half-reaction with the higher oxidation potential will occur at the anode.
From standard electrode potential tables, we can see that the reduction potential for the half-reaction involving water, 2H2O (l) + 2e^- -> H2 (g) + 2OH^- (aq), is higher than the reduction potential for the half-reaction involving Br^-, 2Br^- (aq) -> Br2 (g) + 2e^-. Therefore, at the cathode, H2 gas will be produced.
Similarly, the oxidation potential for the half-reaction involving Br^-, 2Br^- (aq) -> Br2 (g) + 2e^-, is higher than the oxidation potential for the half-reaction involving water, 2H2O (l) + 2e^- -> H2 (g) + 2OH^- (aq). Therefore, at the anode, Br2 gas will be produced.
For AgF(aq):
- Oxidation half-reaction: 2F^- (aq) -> F2 (g) + 2e^-
- Reduction half-reaction: Ag^+ (aq) + e^- -> Ag (s)
The half-reaction with the higher reduction potential, Ag^+ (aq) + e^- -> Ag (s), will occur at the cathode, and silver metal (Ag) will be produced.
The oxidation half-reaction involving fluoride, 2F^- (aq) -> F2 (g) + 2e^-, will occur at the anode, and fluorine gas (F2) will be produced.
In summary:
- At the cathode, during the electrolysis of NaBr(aq), hydrogen gas (H2) is produced.
- At the anode, during the electrolysis of NaBr(aq), bromine gas (Br2) is produced.
- At the cathode, during the electrolysis of AgF(aq), silver metal (Ag) is produced.
- At the anode, during the electrolysis of AgF(aq), fluorine gas (F2) is produced.
At the anode, the oxidation of bromide ions (1.07 ) occurs because it has a higher (less negative) potential than the oxidation of water (1.23 ) or fluoride ions (2.87 ). The product of the oxidation of bromide ions is liquid bromine.
At the cathode, the reduction of silver ions (+0.80 ) occurs because it has a higher potential than the reduction of water (0.83 ) or sodium ions (2.71 ). The product of the reduction of silver ions is solid silver