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1. If 1M of ZnCl2 were exposed to a 9.65A for 1000s, how many grams of Zn would plate onto cathode?
a) q = 9.65 x 1000s = 9650C
b) 9650C (1mol e-/ 96500C)(1 mole Zn / 2 mol e-)(65.38g/mol) = 3.3g Zn
I have a question on the conversion factor. Does The 2 mole electron come from
ZnCl2 (in which the oxidation state of Zn = +2), or does it come from the difference in oxidation states between ZnCl2 and Zn (2 - 0 = 2)?
a) q = 9.65 x 1000s = 9650C
b) 9650C (1mol e-/ 96500C)(1 mole Zn / 2 mol e-)(65.38g/mol) = 3.3g Zn
I have a question on the conversion factor. Does The 2 mole electron come from
ZnCl2 (in which the oxidation state of Zn = +2), or does it come from the difference in oxidation states between ZnCl2 and Zn (2 - 0 = 2)?
