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- Redox and Electrochemistry

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Get Started Now- Intro Lesson: a3:12
- Intro Lesson: b9:21
- Intro Lesson: c9:10
- Lesson: 1a7:26
- Lesson: 1b7:43
- Lesson: 1c8:22

In this lesson, we will learn:

- To recall the two methods for completing full redox equations.
- How to complete redox equations using oxidation numbers.
- How to complete redox equations using half-equations.

- We have seen
__two ways to work with redox reactions__:- We can split the redox reaction in half to show the reduction and oxidation processes separately. This gives us two
*half-equations*.

These half-equations show the electrons lost or gained in the separate processes, and the number of electrons must match in the overall equation. - We can assign
*oxidation numbers*to the atoms involved in the reaction. Some oxidation numbers will change from reactants to products as atoms are either reduced or oxidized. The overall change in oxidation numbers must be zero in the overall equation to balance out.

- We can split the redox reaction in half to show the reduction and oxidation processes separately. This gives us two
- This will depend on your curriculum, but generally
__to start a redox question, you will only be given compounds containing major atoms__.

In practice, this means__an unbalanced equation with no H__^{+}__or H__. You must complete the redox equation from this point. In this lesson we will practice completing and balancing them._{2}O present __Worked example using half-equations__:Cr(OH) _{3}+ Cl_{2}$\enspace$→$\enspace$ CrO_{4}^{2-}+ Cl^{-}

This equation needs to be split into its half equations.__This is done by separating the species with major atoms__into separate equations:$\quad$Equation 1: $\quad \qquad \qquad \quad$ Cr(OH) _{3}$\enspace$→$\enspace$ CrO_{4}^{2-}$\qquad \qquad \qquad \qquad \qquad \qquad$Equation 2: $\quad \qquad \qquad \qquad \quad \;$ Cl _{2}$\enspace$→$\enspace$ Cl^{-}$\qquad \qquad \qquad \qquad \qquad \qquad$

First for equation 1, balance the major atoms. The Cr is already balanced so no change:Cr(OH) _{3}$\enspace$→$\enspace$CrO_{4}^{2-}

Then equation 2 where the Cl atoms need to be balanced:Cl _{2}$\enspace$→$\enspace$2CL^{-}

Next, balance the oxygen atoms:Equation 1: $\quad \qquad \qquad \,$ Cr(OH) _{3}+ H_{2}O$\enspace$→$\enspace$ CrO_{4}^{2-}$\qquad \qquad \qquad \qquad \qquad \qquad$Equation 2: $\quad \qquad \qquad \qquad \quad \;$ Cl _{2}$\enspace$→$\enspace$ 2Cl^{-}$\qquad \qquad \qquad \qquad \qquad \qquad \qquad$

Next, balance the hydrogen atoms:$\enspace \:$Equation 1: $\quad \qquad \enspace$ Cr(OH) _{3}+ H_{2}O $\enspace$→$\enspace$ CrO_{4}^{2-}+ 5 H^{+}$\qquad \qquad \qquad \qquad \qquad \qquad$Equation 2: $\quad \qquad \qquad \qquad \quad \;$ Cl _{2}$\enspace$→$\enspace$ 2Cl^{-}$\qquad \qquad \qquad \qquad \qquad \qquad \qquad$

Finally, balance the charge:Equation 1: $\quad \qquad \qquad$ Cr(OH) _{3}+ H_{2}O $\enspace$→$\enspace$ CrO_{4}^{2-}+ 5 H^{+}+ 3e^{-}$\qquad \qquad \qquad$$\qquad$Equation 2: $\quad \qquad \qquad \qquad \quad \enspace \,$ 2e ^{-}+ Cl_{2}$\enspace$→$\enspace$ 2Cl^{-}$\qquad \qquad \qquad \qquad \qquad \qquad$

Now with two complete half-equations, we need to re-combine them__with an equal number of electrons on both sides__. Three are transferred in equation 1, only two are transferred in equation 2, so if we multiply both equations to have six on both sides we will balance:Equation 1: $\quad \qquad \qquad$ 2 x [ Cr(OH) _{3}+ H_{2}O$\enspace$→$\enspace$CrO_{4}^{2-}+ 5H^{+}+ 3e^{-}] $\qquad \qquad \qquad$$\qquad$Equation 2: $\quad \qquad \qquad \qquad \quad \enspace \,$ 3 x [ 2e ^{-}+ Cl_{2}$\enspace$→$\enspace$2Cl^{-}] $\qquad \qquad \qquad \qquad \qquad \qquad$

Multiplying and combining these equations with the electrons removed gives us__a final balanced redox equation__. This is the final equation; atoms and charge must balance now!Full equation: $\quad \qquad \qquad$ 2Cr(OH) _{3}+ 2H_{2}O + 3Cl_{2}$\enspace$→$\enspace$2CrO_{4}^{2-}+ 10H^{+}+ 6Cl^{-}$\qquad \qquad \qquad$__Worked example using oxidation numbers__:SO _{3}^{2-}+ Cr_{2}O_{7}^{2-}$\enspace$→$\enspace$SO_{4}^{2-}+ Cr^{3+}

The total change in oxidation number must be zero. First, calculate oxidation numbers of the major atoms in the reactants and products.

Reactants:

Sulfur in SO_{3}^{2-}, oxidation state +4 (-6 due to oxygen, +4 in sulfur gives 2- overall charge).

Chromium in Cr_{2}O_{7}^{2-}, Oxidation state +6 (-14 due to oxygen, +6 for two Cr atoms gives 2- overall charge).

Products:

Sulfur in SO_{4}^{2-}, oxidation state +6 (-8 due to oxygen, +6 in sulfur gives 2- overall charge).

Chromium in Cr^{3+}ion, oxidation state +3.

$\Delta$OS for sulfur = +2 (from +4 to +6)

$\Delta$OS for chromium = -3 x 2 Cr atoms (from +6 to +3) = -6

$\Delta$OS must balance to zero, therefore:

3 x SO_{3}^{2-}$\enspace$→$\enspace$SO_{4}^{2-}gives total $\Delta$OS of +6

1 x Cr_{2}O_{7}^{2-}$\enspace$→$\enspace$Cr^{3+}gives total $\Delta$OS of -6

This balances the oxidation state.

Balance major atoms:3SO _{3}^{2-}+ Cr_{2}O_{7}^{2-}$\enspace$→$\enspace$3SO_{4}^{2-}+ 2Cr^{3+}

Now, balance oxygen with H_{2}O:3SO _{3}^{2-}+ Cr_{2}O_{7}^{2-}$\enspace$→$\enspace$3SO_{4}^{2-}+ 2Cr^{3+}+ 4H_{2}O

Next, balance hydrogen with H^{+}:3SO _{3}^{2-}+ Cr_{2}O_{7}^{2-}+ 8H^{+}$\enspace$→$\enspace$3SO_{4}^{2-}+ 2Cr^{3+}+ 4H_{2}O

Check that the charge is balanced:3SO _{3}^{2-}+ Cr_{2}O_{7}^{2-}+ 8H^{+}$\enspace$→$\enspace$3SO_{4}^{2-}+ 2Cr^{3+}+ 4H_{2}O

- Introduction
__Balancing redox equations__a)Two methods to balance redox.b)Balancing redox using half-equations.c)Balancing redox using oxidation number. - 1.
**Balance and complete the following redox equations using either oxidation numbers or half-equations. All are in acidic conditions.**a)Sn^{2+}+ IO_{3}^{-}$\enspace$→$\enspace$Sn^{4+}+ I^{-}b)Cr^{3+}+ BiO_{3}^{-}$\enspace$→$\enspace$Bi^{2+}+ Cr_{2}O_{7}^{2-}c)MnO_{4}^{-}+ SO_{2}$\enspace$→$\enspace$Mn^{2+}+ HSO_{4}^{-}