Answer:
2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)
Explanation:
Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.
E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V
The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.
Answer: The concentrations of 
at equilibrium is 0.023 M
Explanation:
Moles of = 
Volume of solution = 1 L
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)
Yes it is available. It will continue catalyzing the reactions until it becomes completely consumed. That's how enzymes work. They work and are eventually consumed in the process completely without altering the reaction in any way other than speeding it up.
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