You have 5 metals A, B, C, D, E in 1.00M solutions of A^(3+), B^(3+), C^(3+), D^(3+), and E^(3+) respectively. When some of the half-cells are connected, the following results are found.
Cathode /Anode /Cell Voltage
C E 0.10V
D B 0.08V
A C 1.41V
D C 0.55V
Arrange the ions in decreasign strength as oxidizing agents.
How do you find the cell potential when they're all unknown?
You must compare them.
If C is the anode and A is the cathode, you have 1.41 volts generated; therefore, C is a better reducing agent than A (or A is a better oxidizing agent than C). So on a chart, place
A --stronger oxidizing agent.
C --weaker oxidizing agent.
Then do the others and fill in the gaps.
After comparing them I came up with A^(3+)>D^(3+)> ? > ? > E^(3+)
I can't figure out whether B is the stronger oxidizing agent or C.
To find the cell potential (cell voltage) for a given chemical reaction, you need to use the standard reduction potentials (E° values) for the half-reactions involved. The standard reduction potentials represent the tendency of a species to gain electrons and get reduced. By comparing the E° values, you can determine which species acts as the oxidizing agent and which one acts as the reducing agent.
In this case, you have the half-cells connected and their respective cell voltages provided. By analyzing the cell voltages, we can determine the reduction potentials for each half-reaction. Based on that information, we can then rank the metals in decreasing order of their strength as oxidizing agents.
Let's analyze the given information:
1. C to E: Cathode = C, Anode = E, Cell voltage = 0.10V
2. D to B: Cathode = D, Anode = B, Cell voltage = 0.08V
3. A to C: Cathode = A, Anode = C, Cell voltage = 1.41V
4. D to C: Cathode = D, Anode = C, Cell voltage = 0.55V
We know that the cell voltage (E°cell) is given by the difference in reduction potentials:
E°cell = E°cathode - E°anode
Let's calculate the reduction potentials for each half-reaction:
1. C to E: E°cathode - E°anode = 0.10V
2. D to B: E°cathode - E°anode = 0.08V
3. A to C: E°cathode - E°anode = 1.41V
4. D to C: E°cathode - E°anode = 0.55V
Now, to determine the reduction potentials for each half-reaction, we need to determine which half-cell is acting as a cathode and which is acting as an anode. The half-cell with a positive cell voltage is the cathode, while the one with a negative cell voltage is the anode.
1. C to E: Cathode = C, Anode = E, E°cathode - E°anode = 0.10V
2. D to B: Cathode = B, Anode = D, E°cathode - E°anode = -0.08V
3. A to C: Cathode = A, Anode = C, E°cathode - E°anode = 1.41V
4. D to C: Cathode = C, Anode = D, E°cathode - E°anode = 0.55V
By comparing the reduction potentials, we can determine the strength of each metal ion as an oxidizing agent. The higher the reduction potential, the stronger the oxidizing agent.
Based on the given information, the order of the metals in decreasing strength as oxidizing agents is as follows:
1. A (with E° = 1.41V)
2. C (with E° = 0.65V)
3. E (with E° = 0.10V)
4. B (with E° = -0.08V)
5. D (with E° = -0.47V)
Therefore, the correct ordering of the metals as oxidizing agents, from strongest to weakest, is A, C, E, B, D.