Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)$ (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)

Step by step solution

01

Balance the Chromium atoms

We have 2 Cr atoms on the left side and 1 Cr atom on the right side. To balance them, multiply the right side by 2: \[Cr_2O_7^{2-}(aq) \longrightarrow 2Cr^{3+}(aq)\]

02

Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

03

Balance Oxygen atoms

There are 7 oxygen atoms on the left and none on the right. Add 7 H2O molecules on the right to balance the oxygen atoms: \[Cr_2O_7^{2-}(aq) \longrightarrow 2Cr^{3+}(aq) + 7H_2O(l)\]

04

Balance charges

The total charge on the left side is -2, and the total charge on the right side is +6. To balance the charges, add 6 electrons (e-) to the right side: \[\underline{Reduction:} \quad Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \longrightarrow 2Cr^{3+}(aq) + 7H_2O(l)\] (b) \(Mn^{2+}(aq) \longrightarrow MnO_4^{-}(aq)\) (acidic solution)

05

Balance the Manganese atoms

Manganese atoms are already balanced (1 Mn atom on each side).

06

Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

07

Balance Oxygen atoms

There are 4 oxygen atoms on the right side and none on the left. Add 4 H2O molecules on the left to balance the oxygen atoms: \[Mn^{2+}(aq) + 4H_2O(l) \longrightarrow MnO_4^{-}(aq)\]

08

Balance charges

The total charge on the left side is +2, and the total charge on the right side is -1. To balance the charges, add 3 electrons (e-) to the right side: \[\underline{Oxidation:} \quad Mn^{2+}(aq) + 4H_2O(l) \longrightarrow MnO_4^{-}(aq) + 8H^+(aq) + 5e^-\] (c) \(I_2(s) \longrightarrow IO_3^{-}(aq)\) (acidic solution)

09

Balance the Iodine atoms

We have 2 I atoms on the left side and 1 I atom on the right side. To balance them, multiply the right side by 2: \[I_2(s) \longrightarrow 2IO_3^{-}(aq)\]

10

Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

11

Balance Oxygen atoms

There are 6 oxygen atoms on the right side and none on the left. Add 6 H2O molecules on the left to balance the oxygen atoms: \[I_2(s) + 6H_2O(l) \longrightarrow 2IO_3^{-}(aq)\]

12

Balance charges

The total charge on the left side is 0, and the total charge on the right side is -2. To balance the charges, add 2 electrons (e-) to the left side: \[\underline{Reduction:} \quad I_2(s) + 6H_2O(l) + 2e^- \longrightarrow 2IO_3^{-}(aq) +12H^+(aq)\] Continue with the remaining half-reactions in a similar manner.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!