Posted by Amber on Saturday, May 28, 2011 at 10:30pm.
Draw a line, something like this, at the bottom of the equation connecting Ag and AgNO3.
|_______________|. Write zero under Ag and +1 under Ag of AgNO3.
Now draw another line upside down from the one above and place it above the equation connecting N of HNO3 and N of NO2. |--------------| N for HNO3 is +5 and N of NO is +2. Write those in similar to the way you did the Ag and AgNO3.That gives you everything you need to balance it. On the Ag to AgNO3, write
|____loss of 1e______|
On the N to N write
|-----gain of 3e ----|.
You must keep the electrons equal; therefore, multiply the Ag to AgNO3 by 3 and the HNO3 to NO by 1. That gives you the following
3Ag + HNO3 ==> 3AgNO3 + NO + H2O
I'm sure you have all of this already, probably a dozen or more times. The next part is the tricky part.
The redox part is balanced. Now you must recognize that the NO3 in AgNO3 is not reduced like the NO3^- to NO. So we need 1 HNO3 for the NO part of the equation(the redox part) and we need another 3 HNO3 to take care of the 3AgNO3 (from the non-redox part). So 3+1 = 4 and we make it 4HNO3 like this.
3Ag + 4HNO3 ==> 3AgNO3 + NO + H2O.
All you need to do now is to balance the H2O (it's 2H2O) and you'll have it. :-)
OMG Doctor Bob you are a huge help!
And you right i did have that a dozen times lol but i wasn't sure if i had it right because i couldn't het the last part to work out for me!
But i have it now so thanks a bunch!