One kind of battery used in watches contains mercury(II) oxide. As current flows, the mercury oxide is reduced to mercury.
HgO(s) + H2O(l) + 2e^- ==> Hg(l) + 2 OH^-(aq)
If 2.5 x 10^-5 amperes flows continuously for 1095 days, what mass of Hg(l) is produced?
How have you approached this problem? You know to calculate coulombs, change to Faraday. etc.
To find the mass of Hg(l) produced, we need to use Faraday's law of electrolysis, which states that the mass of a substance produced or consumed at an electrode is directly proportional to the amount of electricity used. The equation we will use is:
mass = (current × time × molar mass of Hg(l)) / (number of electrons × Faraday's constant)
First, let's identify the given values:
- Current: 2.5 x 10^-5 amperes
- Time: 1095 days (which we will convert to seconds later)
- Number of electrons: From the given balanced equation, we can see that 2 electrons are required to reduce 1 molecule of HgO to Hg(l).
- Molar mass of Hg(l): You would need to look up the molar mass of mercury to find this value.
- Faraday's constant: This constant represents the charge of one mole of electrons and is equal to 96,485 coulombs per mole of electrons.
Now, let's calculate the mass of Hg(l) produced.
Step 1: Convert time to seconds
There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, we can calculate the total time in seconds.
1095 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute = Total time in seconds
Step 2: Calculate the mass using Faraday's law of electrolysis
Using the equation mentioned earlier, we can substitute the given values into the formula:
mass = (current × time × molar mass of Hg(l)) / (number of electrons × Faraday's constant)
Substitute the values into the equation, considering the appropriate units:
mass = (2.5 x 10^-5 A × Total time in seconds × molar mass of Hg(l)) / (2 × 96,485 C/mol)
Step 3: Calculate the final mass
Once you have substituted the values, calculate the final mass.
mass = Result from Step 2
By following these steps and substituting the appropriate values, you can find the mass of Hg(l) produced when 2.5 x 10^-5 amperes flow continuously for 1095 days.