Question

please help! I really do not understand half cells. A concentration cell is prepared by adding 90mL of 0.0100M Cu(NO3)2 and a Cu bar in each half cell.

Answer 1

What is the Ecell of that concentration cell at this point?
b) Guven that Cu(OH)2 is insiluble in water, whaat would be the Ecell be after adding 10mL of o.o15M NaOH to one of the half-cells?

Answer 2

*0.0015M

Answer 3

Ecell= (-0.0592/n )* (log [sol diluted Anode]/[sol concentrated cathode])

Answer 4

for part a is it