queS: an electrochemical cell was constructed using Mg anode and Pt cathode. the cell potential was measured to be 3.8V. Which would be the suitable electrolyte for cathode half cell?
a)HCl
b)Mg(NO3)2
c)PtI2 platinum iodide
d) KNO3
not sure why its A
If the anode was Mg metal then the anolyte must have been Mg
2+.
Oxidation occurs at the anode. Magnesium metal is converted into magnesium ions. According to the data sheet, this has an E
o value of 2.36V
So all we have to do is subtraction.
3.8-2.36=1.44
Which is approximately equal to 1.36V which is the value of the chlorine gas/chloride ion reduction's E
o value.
______________
Alternatively, this can be done through process of elimination.
The cathode is Pt. Pt is an inert metal, and used whenever there are gases involved in the equation. So recalling that chlorine gas gets reduced into chloride ions, A has to be correct.
B is wrong because Mg
2+ is the anolyte, not the catholyte.
C is the weird one. If the cathode is Pt and the catholyte is PtI
2, then platinum must be taking part in the reaction. Logically, this is fine because magnesium is going to be heaps more reactive than platinum. However that's the thing; platinum is so unreactive to the point this probably just won't occur at all. Also, it was probably intended to be a dud option as Pt does not appear on the standard reduction potentials on the data sheet
D is wrong because K is more reactive than Mg. If the cell was constructed with potassium ions then the magnesium metal would have to be the cathode.