Redox Reactions

Cheatsheet Content

1. Introduction to Redox Reactions Redox reactions (reduction-oxidation reactions) involve the transfer of electrons between chemical species. They are fundamental in chemistry, playing roles in energy production, corrosion, and biological processes. Oxidation: Loss of electrons. Increase in oxidation state. Reduction: Gain of electrons. Decrease in oxidation state. LEO says GER: L ose E lectrons O xidation; G ain E lectrons R eduction. Oxidizing Agent (Oxidant): The species that causes oxidation by accepting electrons. It gets reduced. Reducing Agent (Reductant): The species that causes reduction by donating electrons. It gets oxidized. 2. Assigning Oxidation States Rules for assigning oxidation states (numbers) to atoms in compounds and ions: An atom in its elemental form has an oxidation state of 0 (e.g., $O_2$, $Na$, $Cl_2$). The oxidation state of a monatomic ion is equal to its charge (e.g., $Na^+$ is +1, $O^{2-}$ is -2). Group 1 metals are always +1 in compounds. Group 2 metals are always +2. Fluorine is always -1 in compounds. Other halogens are usually -1, but can be positive with oxygen. Hydrogen is usually +1 in compounds, except when bonded to metals (metal hydrides), where it is -1 (e.g., $NaH$). Oxygen is usually -2 in compounds, except in peroxides ($O_2^{2-}$, where it's -1), superoxides ($O_2^-$, where it's -1/2), and when bonded to fluorine (e.g., $OF_2$, where it's +2). The sum of oxidation states in a neutral compound is 0. The sum of oxidation states in a polyatomic ion is equal to the ion's charge. Example: Assigning Oxidation States $KMnO_4$: $K$: +1 (Group 1) $O$: -2 (usually) $\implies 4 \times (-2) = -8$ Let $Mn$ be $x$: $+1 + x + (-8) = 0 \implies x = +7$ $Cr_2O_7^{2-}$: $O$: -2 (usually) $\implies 7 \times (-2) = -14$ Let $Cr$ be $x$: $2x + (-14) = -2 \implies 2x = 12 \implies x = +6$ 3. Balancing Redox Reactions The half-reaction method is commonly used to balance redox reactions: Steps for Acidic Solution: Separate the overall reaction into two half-reactions (oxidation and reduction). Balance all atoms except $O$ and $H$. Balance $O$ atoms by adding $H_2O$ molecules to the side deficient in oxygen. Balance $H$ atoms by adding $H^+$ ions to the side deficient in hydrogen. Balance charge by adding electrons ($e^-$) to the more positive side. Multiply each half-reaction by an integer so that the number of electrons lost in oxidation equals the number gained in reduction. Add the two half-reactions, canceling out electrons and any identical species ($H_2O$, $H^+$) on both sides. Verify that atoms and charges are balanced. Steps for Basic Solution: Follow steps 1-6 for acidic solution. For every $H^+$ ion, add an equal number of $OH^-$ ions to BOTH sides of the equation. Combine $H^+$ and $OH^-$ ions on the same side to form $H_2O$ molecules. Cancel out any redundant $H_2O$ molecules from both sides. Verify that atoms and charges are balanced. Example: Balancing in Acidic Solution $MnO_4^- (aq) + Fe^{2+} (aq) \rightarrow Mn^{2+} (aq) + Fe^{3+} (aq)$ Oxidation Half-Reaction: $Fe^{2+} \rightarrow Fe^{3+}$ Balance atoms (already balanced). Balance charge: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ Reduction Half-Reaction: $MnO_4^- \rightarrow Mn^{2+}$ Balance $Mn$ (already balanced). Balance $O$: $MnO_4^- \rightarrow Mn^{2+} + 4H_2O$ Balance $H$: $8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O$ Balance charge: $8H^+ + MnO_4^- + 5e^- \rightarrow Mn^{2+} + 4H_2O$ Combine: Multiply oxidation HR by 5: $5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$ Add: $8H^+ + MnO_4^- + 5e^- + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} + 5e^-$ Cancel electrons: $8H^+ (aq) + MnO_4^- (aq) + 5Fe^{2+} (aq) \rightarrow Mn^{2+} (aq) + 4H_2O (l) + 5Fe^{3+} (aq)$ 4. Electrochemical Cells Redox reactions are the basis of electrochemical cells, which convert chemical energy into electrical energy (voltaic/galvanic cells) or vice versa (electrolytic cells). Anode: Electrode where oxidation occurs. Site of electron release. (Negative for voltaic, Positive for electrolytic). Cathode: Electrode where reduction occurs. Site of electron gain. (Positive for voltaic, Negative for electrolytic). Salt Bridge: Connects the two half-cells, allowing ion flow to maintain charge neutrality. Electromotive Force (EMF) / Cell Potential ($E_{cell}$): The potential difference between the cathode and anode, driving electron flow. $$E_{cell} = E_{cathode} - E_{anode}$$ or $$E_{cell} = E_{reduction} + E_{oxidation}$$ Standard Electrode Potentials ($E^\circ$): Potentials measured under standard conditions (1 M concentration, 1 atm pressure for gases, 25°C). Nernst Equation: Relates cell potential to concentrations of reactants and products. $$E = E^\circ - \frac{RT}{nF} \ln Q$$ At 25°C: $$E = E^\circ - \frac{0.0592}{n} \log Q$$ where $R$ is the gas constant, $T$ is temperature, $n$ is moles of electrons, $F$ is Faraday's constant, and $Q$ is the reaction quotient. Gibbs Free Energy and $E_{cell}$: $$\Delta G^\circ = -nFE^\circ_{cell}$$ If $\Delta G^\circ 0$, reaction is spontaneous (voltaic cell). If $\Delta G^\circ > 0$, $E^\circ_{cell} 5. Common Applications Batteries: Convert chemical energy into electrical energy via spontaneous redox reactions (e.g., Lead-acid batteries, Lithium-ion batteries). Corrosion: Electrochemical oxidation of metals (e.g., rusting of iron). Electroplating: Using electrolysis to deposit a thin layer of one metal onto another. Biological Processes: Cellular respiration, photosynthesis, and metabolism involve complex redox pathways. Fuel Cells: Convert chemical energy from a fuel (e.g., hydrogen) and an oxidant (e.g., oxygen) into electricity.