in solutie exista ionii Ag⁺ NO3⁻(din AgNO3) si H⁺ OH⁻(din apa)
in timpul electrolizei, au loc procesele:
catod (-): 2Ag⁺+2e----> 2Ag
anod(+): 2OH⁻- 2e----> 1/2O2 + H2O
insumate: 2Ag⁺+2OH⁻---> 2Ag + H2O+1/2O2
2mol Ag⁺ rezulta din 2mol AgNO3---> m= niuxM= 2X170g AgNO3
Volumul de oxigen : V= 1/2Vm= 1/2x22,4 l= 11,2l
2x170gAgNO3.........11,2 L o2
17g...................................x=..........................calculeaza !!!!!
