University Level Notes on Redox Chemistry and Oxidation States and Balancing

Fundamental Definitions of Redox Reactions

Redox reactions consist of two simultaneous processes: oxidation and reduction. These processes can be understood through four distinct criteria: oxygen transfer, hydrogen transfer, electron transfer, and changes in oxidation number.

Oxidation

Oxidation is defined by the following four occurrences:

Gain of Oxygen: An atom or compound gains oxygen atoms. Example: $Na + O_2 \rightarrow Na_2O$ .
Loss of Hydrogen: A compound loses hydrogen atoms. Example: $NH_3 \rightarrow N_2 + H_2$ .
Loss of Electrons: An atom or ion loses one or more electrons, often forming a cation. Example: $Na \rightarrow Na^+ + e^-$ .
Increase in Oxidation Number: The degree of oxidation of an element increases. Example: $SO_2 + \frac{1}{2}O_2 \rightarrow SO_3$ . In this reaction, sulfur increases its oxidation state from+6$.

Reduction

Reduction is defined by the following four occurrences:

Gain of Hydrogen: A compound gains hydrogen atoms. Example:C_2H_4 + H_2 \rightarrow C_2H_6 $.</li><li>Loss of Oxygen: A compound loses oxygen atoms. Example:$ CuO + CO \rightarrow Cu + CO_2 $.</li><li>Gain of Electrons: An atom or ion gains one or more electrons, often forming an anion. Example:$ I_2 + 2e^- \rightarrow 2I^- $.</li><li>Decrease in Oxidation Number: The degree of oxidation of an element decreases. Example:$ MnO_4^- \rightarrow Mn^{2+} $. In this reaction, manganese decreases its oxidation state from$ +2 $.</li></ol><h3 id="4b5eadcb-033c-4508-9ed5-d637233c940b" data-toc-id="4b5eadcb-033c-4508-9ed5-d637233c940b" collapsed="false" seolevelmigrated="true">Oxidation Numbers and Assignment Rules</h3>An oxidation number (or oxidation state) is a numerical value assigned to each atom or ion in a compound to indicate its degree of oxidation. These numbers can be positive, negative, or zero, and the or ign must always be explicitly included.<h4 id="afed52fb-c3f6-4056-bd83-b9d967cb6023" data-toc-id="afed52fb-c3f6-4056-bd83-b9d967cb6023" collapsed="false" seolevelmigrated="true">Rules for Assigning Oxidation Numbers</h4>There are six primary rules for determining oxidation states:<ol><li>Uncombined Elements: Any element in its uncombined state has an oxidation number of zero. This applies to all forms, such as$ S_8 $d$ Zn $.</li><li>Fixed Oxidation Numbers in Compounds:<ul><li>Group 1 Elements: Always possess an oxidation number of$ +1 $.</li><li>Group 2 Elements: Always possess an oxidation number of$ +2 $.</li><li>Fluorine: Always possesses an oxidation number of$ -1 $.</li><li>Hydrogen: Usually possesses an oxidation number of$ +1 $, except in metal hydrides (e.g.,$ NaH $), where it is$ -1 $.</li><li>Oxygen: Usually possesses an oxidation number of$ -2 $, with two major exceptions: in peroxides it is$ -1 $, and in it is$ +2 $.</li></ul></li><li>Monatomic Ions: The oxidation number of an element in a monatomic ion is identical to the ionic charge. For example, the oxidation number for is$ -1 $, and for it is$ +3 $.</li><li>Sum in Compounds: The total sum of all oxidation numbers for every atom in a neutral compound must equal zero.</li><li>Sum in Polyatomic Ions: The total sum of all oxidation numbers for every atom in a polyatomic ion must equal the overall charge of the ion.</li><li>Electronegativity Rule: In any compound or ion, the more electronegative element is assigned the negative oxidation number.</li></ol><h4 id="81005b28-10bf-43b7-b916-a556f8bd3d78" data-toc-id="81005b28-10bf-43b7-b916-a556f8bd3d78" collapsed="false" seolevelmigrated="true">Examples of Finding Unknown Oxidation Numbers</h4><ul><li>$ K_2Cr_2O_7 $: To find the oxidation state of$ Cr $, set up the equation:$ (+1 \times 2) + 2x + (-2 \times 7) = 0 $. Solving for gives$ +6 $.</li><li>$ MnO_4^- $: To find the oxidation state of$ Mn $, set up the equation:$ (x) + (-2 \times 4) = -1 $. Solving for gives$ +7 $.</li><li>$ NaClO_3 $: To find the oxidation state of$ Cl $, set up the equation:$ (+1) + (x) + (-2 \times 3) = 0 $. Solving for gives$ +5 $.</li><li>$ PO_4^{3-} $: To find the oxidation state of$ P $, set up the equation:$ (x) + (-2 \times 4) = -3 $. Solving for gives$ +5 $.</li></ul><h3 id="af9c6de4-27c3-4679-ad47-6fafcccdf339" data-toc-id="af9c6de4-27c3-4679-ad47-6fafcccdf339" collapsed="false" seolevelmigrated="true">Oxidising and Reducing Agents</h3>Since oxidation and reduction occur simultaneously, every redox reaction must contain both an oxidising agent and a reducing agent.<h4 id="207c96a1-f329-43d9-a60b-53f124ea883b" data-toc-id="207c96a1-f329-43d9-a60b-53f124ea883b" collapsed="false" seolevelmigrated="true">Oxidising Agents (Oxidants)</h4>An oxidising agent is a substance that brings about oxidation in another species. It does this by removing electrons from another atom or ion. Consequently:<ul><li>The oxidising agent increases the oxidation number of the other species.</li><li>The oxidation number of the oxidising agent itself decreases during the reaction.</li><li>Typical Examples: Oxygen ($ O_2 $), Chlorine ($ Cl_2 $), and Potassium manganate(VII) ($ KMnO_4 $).</li></ul><h4 id="775ceb35-5c5f-4cc9-9929-b03b9a7397d8" data-toc-id="775ceb35-5c5f-4cc9-9929-b03b9a7397d8" collapsed="false" seolevelmigrated="true">Reducing Agents (Reductants)</h4>A reducing agent is a substance that brings about reduction in another species. It does this by donating (giving) electrons to another atom or ion. Consequently:<ul><li>The reducing agent decreases the oxidation number of the other species.</li><li>The oxidation number of the reducing agent itself increases during the reaction.</li><li>Typical Examples: Hydrogen ($ H_2 $), Potassium iodide ($ KI $), and reactive metals like Aluminium ($ Al $).</li></ul><h4 id="5323a6a8-7da9-423b-a217-68e823713764" data-toc-id="5323a6a8-7da9-423b-a217-68e823713764" collapsed="false" seolevelmigrated="true">Variation in Agent Strength</h4>The capacity of a substance to act as an oxidising or reducing agent depends on its standard electrode potential. Some substances can function as either an oxidising or a reducing agent depending on the reaction conditions and the other reactants. For example, hydrogen peroxide ($ H_2O_2 $) is capable of acting as both an oxidant and a reductant.<h3 id="dae9563b-05c6-4130-b282-7af04e8fcd40" data-toc-id="dae9563b-05c6-4130-b282-7af04e8fcd40" collapsed="false" seolevelmigrated="true">IUPAC Naming and Oxidation States</h3>Systematic naming of chemical compounds utilizes Roman numerals to specify the oxidation state of elements that can exhibit multiple oxidation states.<h4 id="2180a643-b3ec-4bfe-9715-a5360ea46a8e" data-toc-id="2180a643-b3ec-4bfe-9715-a5360ea46a8e" collapsed="false" seolevelmigrated="true">Naming Examples with Iron and Nitrogen</h4><ul><li>Iron Chlorides:<ul><li>Iron(II) chloride: Contains ions; formula is$ FeCl_2 $.</li><li>Iron(III) chloride: Contains$ Fe^{3+} $ions; formula is$ FeCl_3 $.</li></ul></li><li>Oxides of Nitrogen:<ul><li>Nitrogen(I) oxide ($ N_2O $): Nitrogen oxidation state is$ +1 $.</li><li>Nitrogen(II) oxide ($ NO $): Nitrogen oxidation state is$ +2 $.</li><li>Nitrogen(IV) oxide ($ NO_2 $): Nitrogen oxidation state is$ +4 $.</li></ul></li><li>Nitrate Ions and Compounds:<ul><li>Sodium nitrate(III) ($ NaNO_2 $): Contains the ion where Nitrogen is$ +3 $.</li><li>Sodium nitrate(V) ($ NaNO_3 $): Contains the$ +5 $.</li></ul></li></ul><h4 id="10cd6bd2-adcc-48ba-812d-a201a8dca9b4" data-toc-id="10cd6bd2-adcc-48ba-812d-a201a8dca9b4" collapsed="false" seolevelmigrated="true">Specific Naming Nuances</h4><ul><li>Chloric Acids:$ HClO_4 $is called chloric(VII) acid because the chlorine atom has an oxidation state of$ +7 $.</li><li>Omission of Oxidation Numbers: For metal ions with only one possible oxidation state (like magnesium), the oxidation number is often omitted. For example, it is simply magnesium nitrate, and it is potassium sulfate. Salts are also typically named without the oxidation number of the non-metal ion if they are standard forms.</li></ul><h3 id="b2ffdc3d-3907-49dd-b0c4-eea491699161" data-toc-id="b2ffdc3d-3907-49dd-b0c4-eea491699161" collapsed="false" seolevelmigrated="true">Balancing Equations Using Oxidation Numbers</h3>Balancing complex redox equations requires following a specific sequence of steps based on the conservation of electrons and charge.<h4 id="4f5de965-d65e-4d60-902d-771b2eb27795" data-toc-id="4f5de965-d65e-4d60-902d-771b2eb27795" collapsed="false" seolevelmigrated="true">Summary of Steps</h4><ol><li>Identify chemical species undergoing oxidation number changes.</li><li>Balance the total oxidation number increases and decreases.</li><li>Balance the overall ionic charges.</li><li>Balance the remaining atoms (usually Hydrogen and Oxygen).</li></ol><h4 id="a6b195d7-4aa0-4874-bc4b-79e12a100520" data-toc-id="a6b195d7-4aa0-4874-bc4b-79e12a100520" collapsed="false" seolevelmigrated="true">Worked Example 1:$ CuO $and$ NH_3 $</h4>Reaction:$ CuO + NH_3 \rightarrow Cu + N_2 + H_2O $<ul><li>Step 1: Identify ox. no. changes.<ul><li>$ Cu $changes from to$ 0 $(Decrease of$ 2 $).</li><li>$ N $changes from to$ 0 $(Increase of$ 3 $).</li></ul></li><li>Step 2: Balance ox. no. changes.<ul><li>To equalize the changes, multiply the change by$ 3 $(Total decrease =$ -6 $) and the change by$ 2 $(Total increase =$ +6 $).</li><li>Coefficient for becomes$ 3 $; coefficient for becomes$ 2 $.</li><li>Coefficient for product becomes$ 3 $;$ N_2 $remains as is because it already contains two nitrogen atoms.</li><li>Equation so far:$ 3CuO + 2NH_3 \rightarrow 3Cu + N_2 + H_2O $</li></ul></li><li>Step 3: Balance atoms.<ul><li>There are hydrogen atoms in$ 2NH_3 $. These are balanced by placing a$ 3 $in front of$ H_2O $.</li><li>Resulting oxygen count:$ 3 $on both sides.</li></ul></li><li>Final Balanced Equation:$ 3CuO + 2NH_3 \rightarrow 3Cu + N_2 + 3H_2O $</li></ul><h4 id="68d0bd72-41c0-4304-84d1-bb645108df34" data-toc-id="68d0bd72-41c0-4304-84d1-bb645108df34" collapsed="false" seolevelmigrated="true">Worked Example 2:$ MnO_4^- $and$ Fe^{2+} $in Acid</h4>Reaction:$ MnO_4^- + Fe^{2+} + H^+ \rightarrow Mn^{2+} + Fe^{3+} + H_2O $<ul><li>Step 1: Identify ox. no. changes.<ul><li>$ Mn $changes from$ +7 $to$ +2 $(Change =$ -5 $).</li><li>$ Fe $changes from to$ +3 $(Change =$ +1 $).</li></ul></li><li>Step 2: Balance ox. no. changes.<ul><li>Multiply by to match the manganese change of$ 5 $.</li><li>Current state:$ MnO_4^- + 5Fe^{2+} + H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + H_2O $</li></ul></li><li>Step 3: Balance charges.<ul><li>Total reactant charge (excluding):</li><li>Total product charge:$ (+2) + (5 \times +3) = +17 $.</li><li>To bridge the gap between and$ +17 $, we need$ 8 H^+ $on the left side.</li><li>Current state:$ MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + H_2O $</li></ul></li><li>Step 4: Balance hydrogen atoms.<ul><li>$ 8 H^+ $on the left are balanced by on the right.</li></ul></li><li>Final Balanced Equation:$ MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O $</li></ul><h3 id="2c59d031-2bae-4d1d-aef7-98912f6f30bd" data-toc-id="2c59d031-2bae-4d1d-aef7-98912f6f30bd" collapsed="false" seolevelmigrated="true">Disproportionation</h3>Disproportionation is a specific type of redox reaction in which the same element is simultaneously oxidised and reduced.<h4 id="f46ad728-a060-4d38-940e-afe2ecf9da59" data-toc-id="f46ad728-a060-4d38-940e-afe2ecf9da59" collapsed="false" seolevelmigrated="true">Example: Chlorine and Sodium Hydroxide</h4>When chlorine is added to hot concentrated aqueous sodium hydroxide, it undergoes disproportionation. The products are chloride ions ($ Cl^- $), chlorate(V) ions ($ ClO_3^- $), and water.Chemical Equation Evolution:<ul><li>Unbalanced:$ Cl_2 + OH^- \rightarrow Cl^- + ClO_3^- + H_2O $</li><li>Balanced Charges:$ 3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + H_2O $</li><li>Final Balanced Form:$ 3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O

Questions & Discussion

Question 4a: Deduce ox. no. changes and state if oxidation or reduction.

i. 2I^- + Br_2 \rightarrow I_2 + 2Br^- $<ul><li>$ I $: From$ -1 $to$ 0 $(Oxidation)</li><li>$ Br $: From$ 0 $to$ -1 $(Reduction)</li></ul></li><li>ii.$ (NH_4)_2Cr_2O_7 \rightarrow N_2 + 4H_2O + Cr_2O_3 $<ul><li>$ N $: From$ -3 $to$ 0 $(Oxidation)</li><li>$ Cr $: From$ +6 $to$ +3 $(Reduction)</li></ul></li><li>iii.$ As_2O_3 + 2I_2 + 2H_2O \rightarrow As_2O_5 + 4H^+ + 4I^- $<ul><li>$ As $: From$ +3 $to$ +5 $(Oxidation)</li><li>$ I $: From$ 0 $to$ -1 $(Reduction)</li></ul></li></ul>Question 4b: Identify agents in parts (i) and (iii).<ul><li>i. Oxidising agent:$ Br_2 $; Reducing agent:$ I^- $.</li><li>iii. Oxidising agent:$ I_2 $; Reducing agent:$ As_2O_3 $.</li></ul>Question 5: Systematic Names.<ul><li>a.$ Na_2SO_3 $: Sodium sulfate(IV)</li><li>b.$ Na_2SO_4 $: Sodium sulfate(VI)</li><li>c.$ Fe(NO_3)_2 $: Iron(II) nitrate(V)</li><li>d.$ Fe(NO_3)_3 $: Iron(III) nitrate(V)</li><li>e.$ FeSO_4 $: Iron(II) sulfate(VI)</li><li>f.$ Cu_2O $: Copper(I) oxide</li><li>g.$ H_2SO_3 $: Sulfuric(IV) acid</li><li>h.$ Mn_2O_7 $: Manganese(VII) oxide</li></ul>Question 6: Formulae.<ul><li>a. Sodium chlorate(I):$ NaClO $</li><li>b. Iron(III) oxide:$ Fe_2O_3 $</li><li>c. Potassium nitrate(III):$ KNO_2 $</li><li>d. Phosphorus(III) chloride:$ PCl_3 $</li></ul>Question 7: Balance via Oxidation Number Method.<ul><li>a.$ H_2SO_4 + 8HI \rightarrow S + 4I_2 + 4H_2O $</li><li>b.$ 2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O $</li><li>c.$ V^{3+} + I_2 + H_2O \rightarrow VO^{2+} + 2I^- + 2H^+ $(Note: Coefficients must balance both masses and charges via the oxidation state method).</li></ul>Question 8: Multiple Choice. Given the equation:<ul><li>Statement B is correct: The oxidation number of each sulfur atom changes from to$ +6 $.</li><li>Other statements are incorrect:$ Cr $changes from to per atom (not or$ +7 $).$ H $remains$ +1$$.