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si hirmet you will be using a copper anode and a zinc cathoc to those in the pos

ID: 1041540 • Letter: S

Question

si hirmet you will be using a copper anode and a zinc cathoc to those in the post-lab a sample You msi experimentally For this pre-lab Assignment you will be preforming similar In this e alaio to t wia nde Thus the Pbi ons vall form instead of Ca tions using a sample set of data, except that the copper Thus the Pb ions will form instead of Cu a amo de is replaced SAMPLE DATA TABLE Initial mass of lead electrode grams Final mass of lead electrode grams Average current Amps Time of current application seconds Trial 1 17.856 17.766 Trial 2 16.594 16.508 0.469 Trial 3 14.998 14.903 0.471 180.0 0.466 180.0 180.0 3. Determine the number of lead atoms lost from the anode in each trial. The electrolysis process uses two electrons to produce one lead ion (Pb). Note: You will have three answers for this question [18 Calculate the number of lead atoms per gram of lead lost at the anode for each trial. The mass lost at the anode is equal to both the mass of lead atoms lost and the mass of lead ions produced (the mass of the electrons is negligible). Note: You will have three answers for this question 4. [18 points) Calculate the number of lead atoms per mole of lead for each trial. Average these three values, and then calculate the percent error and standard de Consider the theoretical value to be 6.02 x 10 atoms/mol. (28 po

Explanation / Answer

3.TRIAL 1

Molar mass of lead = 207.2 g/mol

Mass of lead lost = 17.856 - 17.766 =0.09 g

Moles of lead = mass/molar mass =0.09g/207.2g/mol

=0.0004344 mol

Now, 1 mol = Na(Avogadro number)=6.022×10^23 atoms

0.0004344 mol = 0.0004344×6.022×10^23 atoms

= 2.62 ×10^20 atoms

4. TRIAL 1

Number of lead atoms per gram = atoms/mass of lead

= (2.62 ×10^20)/0.09 g =29.11×10^20atom/g

5.TRIAL 1

Number of atoms per mol = Pb atoms/mol of Pb

= 2.62×10^20 atom/0.0004344 mol =6.03×10^23atom/mol

*Similarly calculation will be done for other trials.

Then take the average of atoms/mol

% error = (actual value/ theoretical value )×100