Video Transcript
The overall voltage for a mercury cell is found to be positive 1.35 volts. Consider the following electrode potential: HgO solid plus H2O liquid plus two
electrons reacts to form Hg liquid plus two OH minus aqueous. 𝐸 equals positive 0.0977 volts. What must the reduction potential of the other electrode in the cell be? Give your answer to four decimal places.
A mercury cell is an example of a primary galvanic cell. The term primary indicates that this cell is designed to be used and then discarded
when it has run flat. It cannot be recharged. A galvanic cell is a type of electrochemical cell that produces electrical energy
from chemical energy using electrons generated by a spontaneous redox reaction. One example which you may be familiar with is the battery.
This question focuses on a mercury battery used in devices, like photography
equipment. Mercury cells are made from zinc and mercury(II) oxide, HgO. Oxidation will occur at the anode, which is zinc, and reduction will occur at the
cathode, which is mercury(II) oxide, mixed with a little graphite. The equation given in the question is the reduction process that mercury(II) oxide
undergoes. Electrons will flow from the anode, the negative electrode, through the electronics,
and back to the cathode, the positive electrode.
A potassium hydroxide solution between the anode and the cathode functions as a salt
bridge. It allows current to pass by completing the circuit but keeps the different chemicals
separate at the terminals. A high or more positive reduction potential value indicates a higher tendency for
reduction and is therefore associated with the cathode.
We know that mercury(II) oxide is the cathode. And so, we can predict that the reduction potential of the zinc electrode, the anode,
will be lower than mercury’s value of positive 0.0977 volts. A low or more negative reduction potential value indicates a low tendency for
reduction and is therefore associated with the anode.
Let’s now calculate zinc’s reduction potential and confirm this. We can use the following equation with these two values to calculate the reduction
potential of the zinc electrode. 𝐸 cell is equal to 𝐸 cathode minus 𝐸 anode. Since the mercury half-cell is the cathode in common mercury batteries, we can
substitute the reduction potential for this electrode here. And we can substitute the cell’s potential for 𝐸 cell. Solving, we get a value of negative 1.2523 volts, which is the value for 𝐸
anode.
Finally, what is the reduction potential of the other electrode to four decimal
places? The answer is negative 1.2523 volts.
