A galvanic cell consists of La^3+/La half-cell and a standard hydrogen electrode. If the La^3+/La half cell functions as the anode and the cell potential is 2.52 V, what is the reduction potential for the La^3+/La half cell?

the choices are -2.52V,-0.84V, +0.84V,+2.52V

If the La/La^+3 is the anode, and you remember that the anode is where oxidation occurs, then the half reaction at the anode must be

La ==> La^+3 + 3electrons .... Ehalfcell
2H^+ + 2e ==> H2...............Ehalfcell
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[If La is the anode, then H2 must be the cathode and gain electrons.]

This is a little tough to do the computer but let me recopy the half cell part below.
La/La^+3 half cell = what voltage
H^+/H2 half cell = 0 voltage
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total cell reaction = total voltage.
Now if the total voltage is 2.52 volts and that is composed of ??volts + 0 volts = 2.52 volts, I give you three guess (and the first two don't count) as to what is ?? volts. It MUST be 2.52 volts since 2.52 + 0 = 2.52.
Here is where you need to be careful to make sure what the question is asking.
If the half cell is
La ==>La^+3 + 3e and that is 2.52 volts, what's next? The question is what is the REDUCTION potential. That is
La^+3 + 3e ==> La which is just the reverse of what the lab experiment was. So reverse the sign to make the oxidation potential of +2.52 volts = to -2.52 for the reduction potential.

Well, well, well, looks like we have a galvanic cell in the house! Let's see if I can put a smile on your face with this answer.

To find the reduction potential for the La^3+/La half-cell, we need to remember that the anode is where oxidation occurs. So, if the La^3+/La half-cell functions as the anode, that means it's losing electrons. Poor thing!

Now, the overall cell potential is given as 2.52 V, which means the difference in potential between the anode and the cathode. Since the La^3+/La half-cell is the anode, its potential should be less than 2.52 V.

So, we can eliminate the choices +2.52V and -2.52V as those are too extreme. Now we are left with -0.84V and +0.84V.

But wait, that's a little tricky! You see, the sign of the potential depends on how we define the reduction half-reaction. So, it could be either -0.84V or +0.84V.

In this case, the reduction potential for the La^3+/La half-cell is actually +0.84V.

So, drum roll, please! The correct answer is +0.84V. I hope that puts a grin on your face!

To determine the reduction potential for the La^3+/La half cell, we need to use the cell potential and the standard reduction potential for the standard hydrogen electrode.

1. The cell potential of the Galvanic cell is given as 2.52 V.
2. The standard reduction potential for the standard hydrogen electrode is defined as 0 V. This is because it is used as a reference electrode, and its potential is assigned as 0 V by convention.

Since the La^3+/La half-cell functions as the anode, it undergoes oxidation. The cell potential represents the difference in potential between the anode and cathode. Therefore, we can set up the following equation:

Cell potential = E(cathode) - E(anode)

We know that E(cathode) is 0 V, so we can rearrange the equation:

E(anode) = E(cathode) - Cell potential
E(anode) = 0 V - 2.52 V

Therefore, the reduction potential for the La^3+/La half-cell is -2.52 V.

To determine the reduction potential for the La^3+/La half-cell, we need to use the given information and the concept of cell potential.

In a galvanic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Given that the La^3+/La half-cell functions as the anode, it means that oxidation takes place at this half-cell.

The cell potential, given as 2.52 V, represents the difference in potential between the anode and the cathode. It is given by the equation:

Cell potential (Ecell) = E(cathode) - E(anode)

Since the standard hydrogen electrode is the cathode, the reduction potential (E(cathode)) for the standard hydrogen electrode is zero (0 V) by definition. Therefore, we can rewrite the equation as:

2.52 V = 0 V - E(anode)

Rearranging the equation, we find:

E(anode) = -2.52 V

The negative sign indicates that the potential for the anode (La^3+/La half-cell) is in reverse compared to the standard hydrogen electrode. Thus, the reduction potential for the La^3+/La half-cell is -2.52 V.

Therefore, the correct choice is -2.52 V.