Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

[ce{Cr_2O_7^{2-}(aq) rightarrow Cr^{3+}(aq) } nonumber ]

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[ce{HNO_2 (aq) rightarrow NO_3^{-}(aq)} nonumber ]

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

[ce{Cr_2O_7^{2-}(aq) rightarrow 2Cr^{3+}(aq)} nonumber ]

and

[ce{HNO_2(aq) rightarrow NO_3^{-}(aq)} nonumber ]

Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 (ce{H2O}) molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

[ce{Cr_2O_7^{2-} (aq) rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} nonumber ]

and

[ce{HNO_2(aq) + H_2O(l) rightarrow NO_3^{-}(aq) } nonumber ]

Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

[ce{14H^+(aq) + Cr_2O_7^{2-}(aq) rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} nonumber ]

and

[ce{HNO_2 (aq) + H2O (l) rightarrow 3H^+(aq) + NO_3^{-}(aq)} nonumber ]

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

[ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq) rightarrow 2Cr^{3+}(aq) + 7H_2O(l)} nonumber ]

For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:

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[ce{HNO_2(aq) + H_2O(l) rightarrow 3H^+(aq) + NO_3^{-}(aq) + 2e^{-}} nonumber ]

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:

[ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-} (aq) rightarrow 2Cr^{3+} (aq) + 7H_2O(l).} nonumber ]

and

[ begin{align*} 3 times big[ ce{HNO2 (aq) + H2O(l)} &rightarrow ce{3H^{+}(aq) + NO3^{-} (aq) + 2e^{-}} big] [4pt] ce{3HNO2 (aq) + 3H_2O (l)} &rightarrow ce{9H^{+}(aq) + 3NO_3^{-}(aq) + 6e^{-}} end{align*} nonumber ]

Step 7: Add the reactions and cancel out common terms.

[begin{align*} ce{3HNO_2 (aq) + 3H_2O (l)} &rightarrow ce{9H^+(aq) + 3NO_3^{-}(aq) + 6e^{-} } [4pt] ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq)} &rightarrow ce{2Cr^{3+}(aq) + 7H_2O(l)} [4pt] hline ce{3HNO_2 (aq)} + cancel{ce{3H_2O (l)}} + cancel{6e^{-}} + ce{14H^+(aq) + Cr_2O_7^{2-} (aq)} &rightarrow ce{9H^+(aq) + 3NO_3^{-}(aq)} + cancel{6e^{-}} + ce{2Cr^{3+}(aq)} + cancelto{4}{7}ce{H_2O(l)} end{align*} ]

The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:

[ce{3HNO_2(aq) + 5H^+(aq) + Cr_2O_7^{2-} (aq) rightarrow 3NO_3^{-}(aq) + 2Cr^{3+}(aq) + 4H_2O(l)} nonumber ]

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Category: WHICH