Cathode: "Cr^{3+} + 3e^- \\rightarrow Cr^0"
"m(Cr) = Z\\times I \\times t = \\frac{52.00 g Cr}{3(96500)} \\times \\frac{5.8 C}{s} \\times 2 min \\times \\frac{60 s}{1 min} = 0.125 g"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment