How do I balance this equation?

Question

Ag + Cu(NO3)2 ---> ____________

the 3 and 2 are subscript numbers.

it's a single displacement reaction, and Ag has a charge of 1, NO3 has a charge of -1.

can you explain every single step, please, because i'm really confused.

Thanks!

Answer 1

Ag + Cu(NO3)2 ---> ____________

1) Rewrite everything with its charge
Ag^+1+ Cu^2 NO3^-1---->

2) Positive charge with negative charge, Ag is positive and NO3 is negative. They are together. Cu stands alone

Ag + Cu(NO3)2--->AgNO3+Cu

3) balance
Notice that there are 2 CO3 on the left, so you also need 2 NO3 on the right.
When you put the 2 in front on AgNO3, there are 2 Ag so you need 2 Ag on the left

2Ag+Cu(NO3)2--->2AgCO3+Cu

Answer 2

since its a single displacement reaction, the products are AgNO3 and Cu. [its AgNO3 and not Ag2NO3 or something like that because one Ag's +1 charge is satisfied by one NO3 and thats it]
To balance it, as you see 2 NO3 groups on the left, you can write 2 before AgNO3 on the right. But this makes it 2 Ag on the right. To balance this, write 2 before the Ag on the left. Now, you see that the number of Ag, Cu and NO3 is balanced:
2Ag +Cu(NO3)2 ----> 2AgNO3 + Cu

Since this is a simple equation, this method works fine, but when more complex eqns are involved, the general methodology is to write a,b ,c...before each compound (or element) in the equation and balance the number of a particular element on either side.
For this case,
aAg + bCu(NO3)2 -----> cAgNO3 + dCu.
number (or mole) balance gives:
Ag: a = c
Cu: b = d
N: 2b = c
O: 6b = 3c
The last 2 eqns are the same because N and O appear in a constant proportion as part of 1 group only. This may not always be the case.
Thus, a,b,c,d becomes 2b,b,2b,b which is the same as 2,1,2,1 as the stoichiometric coefficients can all be multiplied or divided by a common factor. check: 2,1,2,1 are the coeff we used to balance in a first straight forward way above.

Answer 3

2Ag + Cu(NO3)2 ==> 2AgNO3 + Cu
Copper takes on (adds) 2 electrons (Cu+2 + 2e ---> Cu)
Silver gives up just one electron (Ag ----> Ag+1 + 1 e)
For the number of electrons in oxidation (loss of electrons) to balance the number of electrons in reduction (gain of electrons) you must multiply the silver half reaction by 2.
2(Ag ----> Ag+1 + 1 e)
(Cu+2 + 2e ---> Cu)
Now add these 2 half reactions together and cancel the electrons.
2Ag + Cu+2 + 2e ===> 2Ag+1 + Cu + 2e

2Ag + Cu+2 ===> 2Ag+1 + Cu
The nitrate ion (NO3) is a spectator ion and does not influence the transfer of electrons, but it should be transferred quantitatively. (same number on each side)

Answer 4

There is no reaction b/c Cu is more reactive than Ag and it is already in aqueous state.

Answer 5

2Ag + Cu(NO3)2 ---> 2AgNO3 + Cu2+