<p>So I picked D.
Princeton Review says it is B. However, I went through the correct process and I don't think I made any mistakes... or did I? In the book, it says to plug in the choices into the equation to see which one works, but they don't take into account the electron transfer. Is this one of PR's Misprints or did I make a mistake?</p>
<p>Okay, just use the PR method. and Like Omarr said don't make it too complex, It's a MC question. Go from Choice C (3) and plug it in, then check if everything balances, and it doesn't so go to choice B (2) and everything balances, more specifically you can look at the Oxygen and that balances to its simplest form. Also, you chose Choice D (4), which is the double of Choice B (2), so make sure you are simplifying your answer.</p>
<p>no no no no no!!!!
Unfortunately you can't take shortcuts and lildude's answer of 4 is correct. (Although dude did forget the 2 in front of the H2O in his/her balanced equation.) This is a mistake in the PR book. If you're not sure whether to believe me, google "nitric acid copper balanced reaction" and you should be able to find at least a couple of answers.
If you google, please note that either NO or NO2 can be formed depending on the conditions of the reaction. If NO is the product, obviously the coefficients change.</p>
<p>BTW ... guess and check isn't always bad, but if you are going to try to balance a redox by guess and check, you also have to check the charge balance. If the coeff. of H+ is only 2, the left side of the equation has a net charge of 1+ at the most (if the coeff of NO3- is 1) and the right side is 2+ at the least (if the coeff of Cu2+ is 1). Therefore the coeff. of H+ must be greater than 2.</p>
<p>Just for reference, the following redox balancing questions were on old AP exams. Granted, there's usually only one of these in the MC section, but now that balancing is required on Question 4, it could show up there as well.</p>
<p>__ Cr2O72¯ + __ e¯ + __ H+ → __ Cr3+ + __ H2O(l)
When the equation for the half reaction above is balanced with the lowest whole-number coefficients, the coefficient for H2O is
(A) 2
(B) 4
(C) 6
(D) 7
(E) 14 </p>
<p>. . . . Ag+ + . . . AsH3(g) + . . . OH¯ → . . . Ag(s) + . . . H3AsO3(aq) + . . . H2O
When the equation above is balanced with lowest whole-number coefficients, the coefficient for OH¯ is
(A) 2
(B) 4
(C) 5
(D) 6
(E) 7</p>
<p>here's how you do it (broken down into simple steps)
let me use gfaith's first ex. above as a guide...</p>
<p>1) Balance out non H or O items
-everything's okay here</p>
<p>2) Balance out O's with H2O's
1 Cr2O72¯ + __ e¯ + __ H+ → 1 Cr3+ + 7 H2O(l)</p>
<p>3) Balance out the H20's with H+'s on the other side
1 Cr2O72¯ + __ e¯ + 14 H+ → 1 Cr3+ + 7 H2O(l)</p>
<p>4) Add up the charges on both sides... then balance out the charges by adding e- onto the side that is more positive...
1 Cr2O72¯ + __ e¯ + 14 H+ → 1 Cr3+ + 7 H2O(l)
13+ on left, 3+ on right ... so 10- will be needed on left 1 Cr2O72¯ + 10 e¯ + 14 H+ → 1 Cr3+ + 7 H2O(l)</p>
<p>This is your final result, but only if it is acidic... if it says it's basic... then there's one more step...</p>
<p>5) add OH- on both sides... just enough to cancel out the H+'s (H+ + OH- = H2O)
1 Cr2O72¯ + 10 e¯ + 14 H+ → 1 Cr3+ + 7 H2O(l)
There are 14 H+'s so must add 14OH-
1 Cr2O72¯ + 10 e¯ + 14 H2O → 1 Cr3+ + 7 H2O(l) + 14OH-</p>
<p>Nicely explained Chopin! Just two comments
3) should say balance the H's (not H2O's) with H+ on the other side (there may be H's in species other than water - they need to be counted too).</p>
<p>And of course, as hopefully everyone knows, if you need to combine two half reactions into a whole redox, insert step (4a):
4a) Multiply each half reaction by a whole number as necessary so electrons cancel when half-reactions are added.</p>