Questions: Drop the coefficients into the boxes to balance the redox equation shown in the figure. While it is conventional to not show coefficients of 1, in this exercise it is okay to do so MnO4, C2O42- + Mn2+ + CO2 + H2O 1 2 4 5 6 8 10 12 16

Drop the coefficients into the boxes to balance the redox equation shown in the figure. While it is conventional to not show coefficients of 1, in this exercise it is okay to do so

MnO4, C2O42- + Mn2+ + CO2 + H2O
1
2
4
5
6
8
10
12
16

Transcript text: Drop the coefficients into the boxes to balance the redox equation shown in the figure. While it is conventional to not show coefficients of 1, in this exercise it is okay to do so Provide your answer below: $\mathrm{MnO}_{4}$ $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}+$ $\mathrm{Mn}^{2+}+$ $\mathrm{CQ}_{2}+$ $\mathrm{H}_{2} \mathrm{O}$ 1 2 4 5 6 8 10 12 16

Solution

Solution Steps

To balance the given redox reaction in neutral conditions, we need to follow a systematic approach. The reaction involves permanganate ion ($\mathrm{MnO}_4^-$) and oxalate ion ($\mathrm{C}_2\mathrm{O}_4^{2-}$).

Step 1: Identify Oxidation and Reduction Half-Reactions

Oxidation Half-Reaction: $\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_2$
Reduction Half-Reaction: $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}$

Step 2: Balance Atoms Other Than Oxygen and Hydrogen

Oxidation Half-Reaction: $\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2$
Reduction Half-Reaction: $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}$

Step 3: Balance Oxygen Atoms by Adding Water

Oxidation Half-Reaction: Already balanced for oxygen.
Reduction Half-Reaction: $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Step 4: Balance Hydrogen Atoms by Adding Protons

Oxidation Half-Reaction: No hydrogen to balance.
Reduction Half-Reaction: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Step 5: Balance Charge by Adding Electrons

Oxidation Half-Reaction: $\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2\mathrm{e}^-$
Reduction Half-Reaction: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Step 6: Equalize the Number of Electrons Transferred

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize the electrons:

Oxidation Half-Reaction: $5\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 10\mathrm{CO}_2 + 10\mathrm{e}^-$
Reduction Half-Reaction: $2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10\mathrm{e}^- \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$

Step 7: Combine the Half-Reactions

Add the balanced half-reactions together:

\[ 2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} + 16\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \]

Step 8: Simplify for Neutral Conditions

Since the reaction is in neutral conditions, the protons ($\mathrm{H}^+$) and water ($\mathrm{H}_2\mathrm{O}$) will cancel out appropriately, leaving:

\[ 2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \]

Final Answer

The balanced redox equation with coefficients is:

\[ \boxed{ \begin{align_} 2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} &\rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \end{align_} } \]

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