I don't really understand these questions and their processes. And I am a bit iffy on redox reactions. For problem 3, can you show a step by step process how to do the problem?
Problem 2, I do not fully understand how to do it at all. How would you know the resulted products?
And for problem 1, can you help me go over it? Reducing and oxidizing agent..I want to understand.
1. Could you produce bromine, Br2, by the electrolysis of a CuBr2(aq) solution?
What are the products of this electrolysis at the anode? At the cathode? (Show your work).
2. Cd^2+(aq) + 2e^- ----> Cd(s)
Eo = 0.40 V
Al(OH)4^-(aq) + 3e^- ----> 4OH^-(aq) + Al(s)
Eo = -2.310 V
A galvanic cell created using the above components will have a standard cell potential, Eocell, of _______ and an overall reaction of:
Which component of the above cell
Is the best reducing agent?
Undergoes oxidation?
Is the best oxidizing agent?
Undergoes reduction?
3. Balance the following reaction in acidic solution:
PbO2(s) + Hg(l) ----> Hg2^2+ + Pb^2+
#1.
Br2 + 2e ==> 2Br^- written as a reduction. (Reduction is the gain of electrons and this gained e). Look in the table of standard reduction potentials (tabulated I suspect in your text). If Br2 + 2e ==> 2Br^- is more less positive than 1.23 (volts for electrolysis of H2O at that electrode), then Br2 will not be produced there. Instead you will obtain 2H2O ==> O2 + 4H^+ + 4e.
At the cathode, Cu^2+ will be attracted and will be plated out if Eo Cu is more positive than zero.
Sure! I can help you with each of these problems one by one.
Problem 1:
To determine if we can produce bromine (Br2) by the electrolysis of a CuBr2(aq) solution, we need to consider the reduction potentials of Cu2+ and Br2.
First, we need to find the half-reactions for each electrode. At the anode (positive electrode), oxidation takes place. The half-reaction is as follows:
Cu(s) --> Cu2+(aq) + 2e-
At the cathode (negative electrode), reduction takes place. The half-reaction for bromine is:
Br2(aq) + 2e- --> 2Br-(aq)
Now we compare the reduction potentials. The more positive reduction potential indicates a stronger reducing agent (easily reduced) and the more negative reduction potential indicates a stronger oxidizing agent (easily oxidized).
The standard reduction potential for Cu2+/Cu is +0.34 V, while for Br2/2Br- is +1.07 V. Since the reduction potential for bromine is higher than that of Cu2+, it means that Cu2+ is a stronger oxidizing agent than Br2. Therefore, bromine cannot be produced by the electrolysis of a CuBr2(aq) solution.
Problem 2:
To determine the standard cell potential (Eocell) and the overall reaction for the galvanic cell containing Cd^2+(aq), Al(OH)4^-(aq), Cd(s), and Al(s), we need to compare the reduction potentials of the half-reactions.
The reduction half-reaction for Cd^2+ is:
Cd^2+(aq) + 2e- --> Cd(s)
Eo = 0.40 V
The reduction half-reaction for Al(OH)4^- is:
Al(OH)4^-(aq) + 3e- --> 4OH-(aq) + Al(s)
Eo = -2.310 V
To find the overall reaction, we need to flip one of the reactions and multiply it by a factor so that the electrons cancel out. In this case, we'll flip the Cd^2+ reaction and multiply it by 3:
3Cd(s) --> 3Cd^2+(aq) + 6e-
Then, add this equation to the Al half-reaction to cancel out the electrons:
3Cd(s) + 3Cd^2+(aq) + 6e- + Al(OH)4^-(aq) + 3e- --> 3Cd^2+(aq) + 3Cd(s) + 6e- + 4OH^-(aq) + Al(s)
Simplifying the equation gives us the overall reaction:
3Cd^2+(aq) + Al(OH)4^-(aq) --> 3Cd(s) + 4OH^-(aq) + Al(s)
The standard cell potential (Eocell) can be calculated by adding the reduction potentials of the half-reactions:
Eocell = Eo(reduction of Cd^2+) - Eo(reduction of Al(OH)4^-)
Eocell = 0.40 V - (-2.310 V)
Eocell = 0.40 V + 2.310 V
Eocell = 2.710 V
So, the standard cell potential (Eocell) of the galvanic cell is 2.710 V.
To determine which component is the best reducing agent (easily oxidized) and which undergoes oxidation, we look at the reduction potentials. The higher the reduction potential, the stronger the reducing agent.
In this case, Cd^2+ has a lower reduction potential than Al(OH)4^-. Therefore, Cd^2+ is the best reducing agent and undergoes oxidation.
Conversely, Al(OH)4^- has a higher reduction potential than Cd^2+. Therefore, Al(OH)4^- is the best oxidizing agent and undergoes reduction.
Problem 3:
To balance the redox reaction in acidic solution between PbO2(s) and Hg(l), we need to follow these steps:
Step 1: Assign oxidation numbers to each element. In this case, PbO2 has an oxidation number of +4 for Pb, and Hg has an oxidation number of 0.
Step 2: Identify the species that undergoes reduction and oxidation. In this case, Pb is being reduced from +4 to +2 (losing electrons), while Hg is being oxidized from 0 to +2 (gaining electrons).
Step 3: Write the half-reactions for reduction and oxidation:
Reduction: PbO2(s) + 4H+(aq) + 2e- → Hg2^2+(aq) + 2H2O(l)
Oxidation: Hg(l) → Hg2^2+(aq) + 2e-
Step 4: Balance the atoms in each half-reaction by adding H2O molecules and H+ ions as needed. In this case, we need to add 4 H+ ions to the reduction half-reaction:
PbO2(s) + 4H+(aq) + 2e- → Hg2^2+(aq) + 2H2O(l)
Step 5: Balance the charges in each half-reaction by adding electrons. In this case, we need to add 2 electrons to the oxidation half-reaction:
Hg(l) → Hg2^2+(aq) + 2e-
Step 6: Multiply each half-reaction by the appropriate factor so that the electrons cancel out when adding the two half-reactions together. In this case, we need to multiply the oxidation half-reaction by 2:
2Hg(l) → Hg2^2+(aq) + 4e-
Now the electrons cancel out, and we can add the two half-reactions together:
PbO2(s) + 4H+(aq) + 2e- + 2Hg(l) → Hg2^2+(aq) + 2H2O(l) + 2Hg2^2+(aq) + 4e-
Simplifying the equation gives us the balanced redox reaction:
PbO2(s) + 4H+(aq) + 2Hg(l) → 2Hg2^2+(aq) + 2H2O(l) + Pb^2+(aq)
So, the balanced redox reaction in acidic solution is:
PbO2(s) + 4H+(aq) + 2Hg(l) → 2Hg2^2+(aq) + 2H2O(l) + Pb^2+(aq)
I hope this helps you understand these problems better! Let me know if you have any more questions.