Using the concepts of 'oxidation numbers' and 'partial ionic equations', balance the following redox reactions which all take place in acid solution. (a) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\) $$ \rightarrow \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} $$ (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+\mathrm{Nal}\) (c) \(\mathrm{KIO}_{3}+\mathrm{KI}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}\) (d) \((\mathrm{COOH})_{2}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\) $$ \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{MnSO}_{4}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$

Step by step solution

01

Title: Identifying the Reduction and Oxidation Half-Reactions for (a)

First, assign oxidation numbers to all atoms in the molecules. Oxygen is assigned a -2 except in peroxides where each oxygen has a -1. The sum of the oxidation numbers for a neutral compound is zero. The atom that is reduced is Cr in \(K_2Cr_2O_7\) (from +6 to +3) and the atom that is oxidized is Fe in \(FeSO_4\) (from +2 to +3). Write out the two half reactions: \(Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} : reduction half-reaction\)\(Fe^{2+} \rightarrow Fe^{3+} : oxidation half-reaction\)

02

Title: Balancing the Half-Reactions for (a)

Next, balance each half reaction. Balancing of half-reactions involves balancing all atoms other than O and H first, then balance O by adding water molecules, H by adding protons and finally, balance the charge by adding electrons as needed on either side. Lastly, they are scaled so that number of electrons lost in the oxidation half reaction equals the number gained in the reduction half reaction. The balanced half reactions are: \(Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\)\(6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-\)

03

Title: Adding the Half-Reactions Together for (a)

To get the final equation, add the two balanced half reactions together and simplify.\(Cr_{2}O_{7}^{2-} + 14H^{+} + 6Fe^{2+} \rightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}\)

04

Title: Balancing B, C, and D

The remaining reactions (b, c, d) are to be balanced following the same procedure. Assign oxidation numbers to atoms in the molecules, write out the half reactions, balance them, and then add them together. The balanced reactions for (b, c, d) are as follows: (b) \(2I^{-} \rightarrow I_{2} + 2e^-\), \(\)\(2S_{2}O_{3}^{2-} + I_{2} \rightarrow S_{4}O_{6}^{2-} + 2I^-\)(c) \(5I^{-} \rightarrow I_{2} +5e^-\), \(\)\(IO_{3}^{-} + 6H^{+} + 5e^- \rightarrow \frac{1}{2} I_{2} + 3H_{2}O \)(d) \(2COOH \rightarrow 2CO_{2} + 2H^{+} + 2e^-\), \(\)\(MnO_{4}^- + 2H^{+} + 2e^- \rightarrow Mn^{2+} + 2H_{2}O\)

05

Final Equations

Finally, for each equation, the ions and molecules are combined with the proper coefficients to balance the charges and species: (a) \( K_{2}Cr_{2}O_{7} + 6FeSO_{4} + 14H_{2}SO_{4} \rightarrow Cr_{2}(SO_{4})_{3} + 3Fe_{2}(SO_{4})_{3} + K_{2}SO_{4} + 7H_{2}O \)(b) \(2Na_{2}S_{2}O_{3} + I_{2} \rightarrow Na_{2}S_{4}O_{6} + 2NaI\)(c) \(6KI + KIO_{3} + 3H_{2}SO_{4} \rightarrow 3K_{2}SO_{4} + 3I_{2} + 3H_{2}O\)(d) \(2(COOH)_{2} + 2KMnO_{4} + 3H_{2}SO_{4} \rightarrow 4K_{2}SO_{4} + 2MnSO_{4} + 4CO_{2} + 7H_{2}O\)

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