Hi, how do you solve this question?
"A student set up the following galvanic cell as shown [zinc anode, copper cathode, ZnSO4 and CuSO4 electrolyte solutions, KCl salt bridge.] Both electrodes were weighed before beginning the experiment and each beaker contained 200mL of a suitable electrolyte at 0.100molL-1 concentration. After a short period of time, the student reweighed the copper electrode and found that it had increased in mass by 0.435g. What was the final concentration of the zinc electrolyte?"
(Answer=0.134molL-1)
Copper cathode oxidised, hence its mass increased as it pulled copper ions out of solution which precipitated on the metal surface. As the cathode gained 0.435g of copper, then 0.435g of copper must have been pulled out of solution. Given the molar mass of copper is
63.55, by
n=mass/Molarmass, n=0.435/63.55,n=0.00684500393... moles. Therefore, the copper cathode gained
0.00684500393... moles of copper.
Cu2+(aq) + 2e- --> Cu(s) and Zn(s) --> Zn2+(aq) + 2e-, which can be written as Cu2+(aq) + Zn(s) --> Cu(s) + Zn2+(aq)
Given that the molar ratio is 1:1, you can determine that the zinc anode lost 0.00684500393... moles of zinc, meaning that the electrolyte solution gained the same moles of zinc ions by the same equation.
Now find the initial moles of Zn2+ in solution, which you can do by
n=cv, n=0.1*0.2 (remembering to convert 200ml to 0.2L because of SI units),
n=0.02 moles.
From here you simply add the initial moles of the solution with the moles which the zinc anode added to its electrolyte solution (
0.02 + 0.00684500393... = 0.02684500393... moles).
Then just solve c=n/v using the new moles of zinc ions in solution and the same volume,
n=0.02684500393.../0.2 = 0.1342250197mol/L.
Hope this helps,
~Nick