How to do basic redox reactions

Step 2: Balance all atoms other than H and O.
Done

Step 3: Balance O by adding H₂O to the deficient side. MnO₄⁻ → MnO₂+ 2H₂O

CN⁻+ H₂O → CNO⁻

Step 4: Balance H by adding H⁺ to the deficient side. MnO₄⁻+ 4H⁺ → MnO₂+ 2H₂O

CN⁻+ H₂O → CNO⁻ + 2H⁺

Step 5: Balance charge by adding electrons to the more positive side. MnO₄⁻+ 4H⁺ + 3e⁻ → MnO₂+ 2H₂O

CN⁻+ H₂O → CNO⁻ + 2H⁺ + 2e⁻

Step 6: Multiply each half-reaction by a factor that gives the lowest common multiple of the electrons transferred. In this case, the lowest common multiple of 3 and 2 is 6.

We multiply the first half-reaction by 2 and the second half-reaction by 3.

2 × [MnO₄⁻+ 4H⁺ + 3e⁻ → MnO₂+ 2H₂O]
3 × [CN⁻+ H₂O → CNO⁻ + 2H⁺ + 2e⁻]

Step 7: Add the two half-reactions, cancelling any like terms.

2MnO₄⁻+ 3CN⁻+ 2H⁺ → 2MnO₂ + 3CNO⁻ + H₂O

This is the balanced equation in acid solution. We must now convert to base solution.

Step 8: Add enough multiples of the equations H⁺ + OH⁻ → H₂O or
H₂O → H⁺ + OH⁻ to cancel the H⁺ in the redox equation, cancelling like terms.

2MnO₄⁻+ 3CN⁻+ 2H⁺ → 2MnO₂ + 3CNO⁻ + H₂O
2H₂O → 2H⁺ + 2OH⁻

2MnO₄⁻+ 3CN⁻+ H₂O → 2MnO₂ + 3CNO⁻ + 2OH⁻

Step 9: Check that atoms balance. On the left: 2 Mn; 9 O; 3 C; 3 N; 2 H

On the right: 2 Mn; 9 O; 3 C; 3 N; 2 H

Step 10: Check that charges balance. On the left: 2- + 3- = 5-

On the right: 3- + 2- = 5-

The balanced equation is
2MnO₄⁻+ 3CN⁻+ H₂O → 2MnO₂ + 3CNO⁻ + 2OH⁻

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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:

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