From the information in the question, the E° and E for the cell is 0.00 V and 0.12 V.
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
We know that E°cell = 0.00 V since the anode and cathode are both made up of cadmium.
Substituting the given values into the Nernst equation;
Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)
Ecell = 0.00 V - 0.0296 log(1 × 10^-4)
Ecell = 0.12 V
Learn more: brainly.com/question/8646601
First you need to calculate the number of moles of aluminium and copper chloride.
number of moles = mass / molecular weight
moles of Al = 512 / 27 = 19 moles
moles of CuCl = 1147 / 99 = 11.6 moles
From the reaction you see that:
if 2 moles of Al will react with 3 moles of CuCl
then 19 moles of Al will react with X moles of CuCl
X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.
Answer:
2 atoms of nitrogen are present.
