AP Chem Question

Test Preparation AP Test Preparation
1

<p>Wondering if any of you could help me balance the following redox reaction (showing work would be appreciated!):</p>

<p>Cr2O7^-2 (aq) + H2S (g) + H+ (aq) yields Cr+3 (aq) + S(s) + H20 (l)</p>

<p>When the equation above is correctly balanced and all coefficients are reduced to lowest whole-number terms, the coefficient for H+ is:</p>

<p>A. 2
B. 4
C. 6
D. 8
E. 14</p>

<p>For some reason, when I run through this, I keep getting E (14) as the answer, but apparently the correct choice is D.</p>

2

<p>That's because you're stopping at the half-reaction for the reduction of Cr2O7(2-) to Cr(3+), which is:</p>

<p>14H(+) + 6e(-) + Cr2O7(2-) ---> 2Cr(3+) + 7H2O</p>

<p>However, you also have to balance the half reaction for the oxidation of H2S to S, then add the two half reactions and balance the overall equation. The half reaction for oxidation of H2S is:</p>

<p>H2S ---> S + 2e(-) + 2H(+)</p>

<p>In order to get the electrons to cancel when adding the two half reactions together, you have to multiply the second reaction (oxidation of H2S) by 3, and then when you add them together, you get:</p>

<p>Cr2O7(2-) + 3H2S + 14H(+) --> 2Cr(3+) + 3S + 6H(+) + 7H2O</p>

<p>A little simple algebra, we subtract 6H(+) from both sides so that H(+) is only present on one side of the reaction, and the final equation is:</p>

<p>Cr2O7(2-) + 3H2S + 8H(+) --> 2Cr(3+) + 3S + 7H2O</p>