Ch_09_summary - CHAPTER 9 0XIDATION AND REDUCTION (TOPICS 9...

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Unformatted text preview: CHAPTER 9 0XIDATION AND REDUCTION (TOPICS 9 AND 19) SUMMARY Basic concepts Oxidation: Loss of Electrons (LEO); Increase in oxidation number (historically: gain of O or loss of H) Reduction: Gain of Electrons; Decrease in oxidation number (GER) (historically: gain of H (or loss of O) Oxidation Number: Represents a charge on a mono-atomic ion or the apparent charge on an element if the more electronegative element is assigned the bonding electrons. Quantitatively, it reflects the oxidation state of an element; an increase in oxidation number by 1 implies a transfer of one e−: Na(s)→Na+ (aq) + e−. An element has an oxidation number = 0. For a mono-atomic ion, oxidation number of element = charge on ion. For a multi-atomic ion, sum of oxidation numbers = charge on ion. Some elements have the same oxidation number in all their compounds (e.g., Na+). Multiple oxidation states are common for d-block elements; the oxidation number is then written in Roman numerals to indicate the oxidation state of the element, e.g.: Fe2+: Fe(II); Fe3+: Fe(III). An oxidizing agent oxidizes another substance; it is reduced in the process (gains electrons); it experiences a decrease in oxidation number; thus, found in its higher/highest oxidation state: e.g.: Cr2O72−; MnO4−; F2. A reducing agent reduces another substance; it is oxidized in the process (loses electrons); it experiences an increase in oxidation number; thus found in its lower/lowest oxidation state: e.g., Na, Ti3+, I−. Reactivity series: A list of elements in order of decreasing reducing ability: An element higher in the series displaces a lower one from its ions: Cu (s) + 2 Ag+ (aq) → 2Ag (s) + Cu2+ (aq). For oxidizing agents, a stronger oxidizing agent displaces a weaker one from its ions: Cl2 (g) + 2 Br− (aq) → Br2 (aq) + 2 Cl− (aq). Electrodes: conductors at which reactions occur in electrochemical and electrolytic cells. Anode: Anions, negative ions, go to anode; electrode at which oxidation occurs (anodic oxidation). Cathode: Cations, positive ions go to cathode; electrode at which reduction occurs (cathodic reduction). A cell contains electrodes placed in an electrolyte solution; the electrodes are connected to an external circuit. An Electrochemical (Galvanic or Voltaic) Cell uses a spontaneous redox reaction to convert chemical energy to electrical energy. It consists of two half-cells: Oxidation occurs at the more negative half-cell. This contains the reducing agent that is oxidized at the (−) electrode, the anode and gives up electrons to the external circuit. Reduction occurs at the more positive half-cell (and thus contains the oxidizing agent that is reduced). The salt bridge maintains electrical neutrality: anions go to the anode, cations go to the cathode. For a current to flow, a potential difference between the two half-cells must exist – this is called the cell potential or e.m.f. of the cell. Cell Diagram: The more negative electrode is written first, to the left, followed by the more positive electrode: Zn(s)/Zn2+(aq) ⏐⏐Cu2+(aq)/Cu(s) © IBID Press 2007 E°cell = E°red + E°ox = (+0.34) + (+0.76) = + 1.10 V. 1 CHAPTER 9 0XIDATION AND REDUCTION (TOPICS 9 AND 19) SUMMARY Electrode potentials The Standard Hydrogen Electrode (SHE) is assigned a value of 0 V under standard conditions of 25°C, solution concentrations of 1.0 mol dm−3 and 1 atmosphere pressure for gases. Positive E°cell indicates a spontaneous reaction (– ΔG°); negative E°cell means a nonspontaneous in the forward direction, or a spontaneous reverse reaction. Standard electrode/reduction potential is the e.m.f. of a half-cell connected to the standard hydrogen potential (at 25°C, 1.0 mol dm−3 solution concentrations and gas at 1 atmosphere pressure). If reduction occurs at the standard half-cell, its sign is +; its sign is negative if reduction occurs at the standard hydrogen electrode (see Table 15). By convention, a cell diagram is written so that the oxidation half-reaction metal/metal ions are placed first, on the left and the reduction half reaction metal ions/metal placed next on the right of the cell diagram; a “⏐” (or “/”) is placed between the metal and its ions and the two aqueous solutions are then separated by a salt bridge represented by two complete or dashed lines: Cell Diagram: Zn(s)/Zn2+(aq) ⏐⏐Cu2+(aq)/Cu(s) E°cell = + 1.10V E°cell is the difference between the two standard electrode potentials. It is common to list: E°cell = E°red (right hand side) – E°red (left hand side) = + 0.34 V – (– 0.76) = +1.10V. It is also possible to determine the cell voltage from the two half oxidation and reduction reactions by using the equation: E°cell = E°red + E°ox. For the above cell, the two half reactions are: Zn(s) → Zn2+(aq) + 2 e–; E°ox.= +0.76V (note: for an electrode, E°ox = – E°red) Cu2+(aq) + 2 e– → Cu(s); E°red = +0.34V E°cell = E°red + E°ox. = +0.76 + 0.34 = +1.10V Electrolysis Electrolysis involves use of electrical energy to carry out a non-spontaneous redox reaction. The electrolyte contains ions that are free to move about (and carry the current: in solution, current involves the movement of charged particles). Cations go to the (−) electrode, the cathode where reduction takes place. Anions go to the (+) electrode, the anode where oxidation takes place. Factors that affect discharge of ions in electrolysis • Position in the electrochemical series: The lower the metal ion is in the electrochemical series, the stronger the oxidizing agent, the more readily it gains electrons (is reduced) at the cathode to form the metal: e.g.: H2 (g) is formed at the cathode instead of Na from NaCl (aq) or NaOH (aq); however, in electrolysis of CuSO4 (aq), Cu is formed, not H2 at (−) electrode (cathode), since Cu is below H in the activity series. • Concentration: Sometimes the rule does not match the observed result, e.g., chloride ions in aqueous solution. For dilute solutions, mainly O2 gas is produced at the (+) electrode (anode) from the oxidation of water. For concentrated NaCl solution, mainly Cl2 gas is produced from the oxidation of Cl− at the (+) electrode. • Nature of the electrode: If an inert electrode is used such as graphite or Pt, it does not take part in the reactions, e.g., in electrolysis of CuSO4 (aq): H2O is oxidized at the (+) electrode to produce O2. But if Cu is used as the (+) electrode, the anode, it is oxidized to Cu2+ ions while Cu2+ ions are reduced to Cu at the (−) electrode and thus [Cu2+] remains the same. © IBID Press 2007 2 CHAPTER 9 0XIDATION AND REDUCTION (TOPICS 9 AND 19) SUMMARY Electrolysis uses direct current from a battery as a source of electrical energy to carry out a non-spontaneous redox reaction. The (+) and (−) terminals of the battery are connected to the electrolytic cell consisting of two electrodes, the (+) electrode, the anode and the (−) electrode, the cathode, in a solution containing cations (M+) and anions (X−). The battery, a source of electrons, supplies electrons to the cathode and removes them from the anode. Cations go to the (−)cathode where reduction takes place to use up the electrons coming in. Anions go to the (+) anode where oxidation occurs to supply the electrons. Applications of Electrolysis Many important metals e.g., Na and Al are prepared by electrolysis of their molten salts; other chemicals, e.g., NaOH are prepared by electrolysis of aqueous solutions e.g. NaCl (aq). Electroplating is a process in which a thin layer of metal is deposited on an electrically conducting surface using electrical energy; it is an important application of electrolytic cells. In copper plating, for example, reduction occurs at (−) cathode: Cu2+(aq) + 2e− → Cu(s); oxidation occurs at (+) anode: H2O(l) →½O2 (g)+2H+(aq)+2e−. The overall reaction: Cu2+ + H2O → Cu + ½ O2 + 2H+. Thus, as the reaction progresses, pH of solution and color intensity (due to Cu2+ (aq)) both decrease. Factors affecting amount of product formed during electrolysis • Charge on the ion: Na+ (l) + 1 e− → Na (l) compared to: Al3+ (l) + 3 e− → Al (l): Thus one mole of Al requires three times the amount of electrons (and thus the charge) compared to 1 mol Na. • Amount of electrons, i.e., the charge passed through the system: Charge (coulombs) = current (amperes) x time (seconds). More current used or charge passed for longer time periods produces more product. (N.B. Shading indicates Ch 19 (AHL) material.) © IBID Press 2007 3 ...
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