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In the first half-reaction, the oxidation number of hydrogen goes from #color(blue)(+1)# on the reactants' side to #color(blue)(0)# on the products' side, so you can say that water is being reduced to hydrogen gas.
#stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) -> stackrel(color(blue)(0))("H")_ (2(g))#
Now, each atom of hydrogen is gaining #1# electron, so you can say that #2# atoms of hydrogen will gain a total of #2# electrons.
#stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) + 2"e"^(-) -> stackrel(color(blue)(0))("H")_ (2(g))#
Now, in order to balance the atoms of oxygen, you need to add water molecules to the side that needs atoms of oxygen and protons, #"H"^(+)#, to the side that needs atoms of hydrogen.
#2"H"_ ((aq))^(+) + stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) + 2"e"^(-) -> stackrel(color(blue)(0))("H")_ (2(g)) + "H"_ 2"O"_ ((l))#
In a basic medium, you must neutralize the protons by adding hydroxide anions, #"OH"^(-)#, to both sides of the equation. The hydroxide anions and the protons will combine to form water.
#overbrace(2"H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-))^(color(red)(= 2"H"_ 2"O"_ ((l)))) + stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) + 2"e"^(-) -> stackrel(color(blue)(0))("H")_ (2(g)) + "H"_ 2"O"_ ((l)) + 2"OH"_ ((aq))^(-)#
This will get you
#2"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) + 2"e"^(-) -> stackrel(color(blue)(0))("H")_ (2(g)) + "H"_ 2"O"_ ((l)) + 2"OH"_ ((aq))^(-)#
which can be simplified to
#2stackrel(color(blue)(+1))("H")_ 2"O"_ ((l)) + 2"e"^(-) -> stackrel(color(blue)(0))("H")_ (2(g)) + 2"OH"_ ((aq))^(-)#
Notice that the half-reaction is balanced in terms of charge because you have
#2 xx 0 + 2 xx (1-) -> 2xx 0 + 2 xx (1-)#
#color(white)(a)#
#color(white)(aaaaaaaaaaaa)/color(white)(a)#
SIDE NOTE: For the second half-reaction, you really need to assume that it's taking place in basic medium because that's really not the case. Methanol can be oxidized to formaldehyde, but the reaction must take place in acidic medium.
In this context, balancing the second half-reaction in a basic medium is not very practical, to put it mildly.
#color(white)(aaaaaaaaaaaa)/color(white)(a)#
In the second half-reaction, the oxidation number of carbon goes from #color(blue)(-2)# on the reactants' side to #color(blue)(0)# on the products' side, so you can say that methanol is being oxidized to formaldehyde.
#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _((aq))#
In this case, each atom of carbon loses #2# electrons, and since you have an atom of carbon on both sides of the half-reaction, you can say that you have
#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _((aq)) + 2"e"^(-)#
The atoms of oxygen are already balanced, but the atoms of hydrogen are not, so add #2# protons on the products' side.
#stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + 2"H"_ ((aq))^(+)#
Once again, the reaction takes place in basic medium, so add hydroxide anions to both sides to neutralize the protons.
#2"OH"_ ((aq))^(-) + stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + overbrace(2"H"_ ((aq))^(+) + 2"OH"_ ((aq))^(-))^(color(red)(=2"H"_ 2"O" _((l))))#
This will get you
#2"OH"_ ((aq))^(-) + stackrel(color(blue)(-2))("C")"H"_ 3"OH"_ ((aq)) -> stackrel(color(blue)(0))("C")"H"_ 2"O" _ ((aq)) + 2"e"^(-) + 2"H"_ 2"O"_ ((l))#
The half-reaction is balanced in terms of charge because you have
#2 xx (1-) + 1 xx 0 -> 1 xx 0 + 2 xx (1-) + 2 xx 0#
Acidic conditions usually implies a solution with an excess of \(\ce{H^{+}}\) concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions. However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add \(\ce{H2O}\) molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons (\(\ce{H^{+}}\)). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.
Example \(\PageIndex{2}\): Balancing in a Acid Solution
Balance the following redox reaction in acidic conditions.
\[\ce{Cr_2O_7^{2-} (aq) + HNO_2 (aq) \rightarrow Cr^{3+}(aq) + NO_3^{-}(aq) } \nonumber\]
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.
\[\ce{Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq) } \nonumber\]
\[\ce{HNO_2 (aq) \rightarrow NO_3^{-}(aq)} \nonumber\]
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:
\[\ce{HNO_2(aq) + H_2O(l) \rightarrow 3H^+(aq) + NO_3^{-}(aq) + 2e^{-}} \nonumber\]
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
\[ \begin{align*} 3 \times \big[ \ce{HNO2 (aq) + H2O(l)} &\rightarrow \ce{3H^{+}(aq) + NO3^{-} (aq) + 2e^{-}} \big] \\[4pt] \ce{3HNO2 (aq) + 3H_2O (l)} &\rightarrow \ce{9H^{+}(aq) + 3NO_3^{-}(aq) + 6e^{-}} \end{align*} \]
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:
\[\ce{3HNO_2(aq) + 5H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 3NO_3^{-}(aq) + 2Cr^{3+}(aq) + 4H_2O(l)} \nonumber\]