Fundamentals of Electrochemistry - W.H. Freeman

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1504T_ch14_270-297 04/06/06 20:12 Page 273 Combining Equations 14-2, 14-3, and 14-4 produces a relation of utmost importance to chemistry: G work E q Relation between free-energy difference and electric potential difference: G nFE (14-5) q nF Equation 14-5 relates the free-energy change of a chemical reaction to the electrical potential difference (that is, the voltage) that can be generated by the reaction. Ohm’s Law Ohm’s law states that current, I, is directly proportional to the potential difference (voltage) across a circuit and inversely proportional to the resistance, R, of the circuit. Ohm’s law: I E (14-6) R Units of resistance are ohms, assigned the Greek symbol (omega). A current of 1 ampere flows through a circuit with a potential difference of 1 volt if the resistance of the circuit is 1 ohm. From Equation 14-6, the unit A is equivalent to V/. The greater the voltage, the more current will flow. The greater the resistance, the less current will flow. Box 14-1 shows measurements of the resistance of single molecules by measuring current and voltage and applying Ohm’s law. Power Power, P, is the work done per unit time. The SI unit of power is J/s, better known as the watt (W). P work s E q s E q s (14-7) Box 14-1 Molecular Wire 3 The electrical conductance of a single molecule suspended between two gold electrodes is known from measurement of voltage and current by applying Ohm’s law. Conductance is 1/resistance, so it has the units 1/ohm siemens (S). To make molecular junctions, a gold scanning tunneling microscope tip was moved in and out of contact with a gold substrate in the presence of a solution containing a test molecule terminated by thiol (ßSH) groups. Thiols spontaneously bind to gold, forming bridges such as that shown here. Nanoampere currents were observed with a potential difference of 0.1 V between the gold surfaces. Au S H + NH 3 N S O The graph shows four observations of conductance as the scanning tunneling microscope tip was pulled away from the Au substrate. Conductance plateaus are observed at multiples of 19 nS. An interpretation is that a single molecule connecting two Au surfaces has a conductance of 19 nS (or a resistance of 50 M). If two molecules form parallel bridges, conductance increases to 38 nS. Three molecules give a conductance of 57 nS. If there are three bridges and the electrodes are pulled apart, one of the bridges breaks and conductance drops to 38 nS. When the second bridge breaks, conductance drops to 19 nS. The exact conductance varies because the environment of each molecule on the Au surface is not identical. A histogram of 7 500 observations in the inset shows peaks at 19, 38, and Au 57 nS. By contrast, the conductance of a chain of Au atoms is 77 S, or 4 000 times greater than 19 nS. Conductance (nS) 76 57 38 19 Data in the chart were obtained at pH 2, at which the amino group is protonated to NH 3 . When the pH was raised to 10, the amino group was neutral. Conductance plateaus became multiples of 9 nS. The conductivity of the molecule changes from 9 nS when it is neutral to 19 nS when there is a positive change on the molecule. The positive charge facilitates electron transfer across the molecule. NH 2 Conductance 9 nS + NH 3 Conductance 19 nS Positive charge in the bridging molecule increases conductivity by a factor of 2. Counts 1 nm 57 38 19 Conductance Change in conductance as Au scanning tunneling microscope tip immersed in dithiol solution is withdrawn from Au substrate. [From X. Xiao, B. Xu, and N. Tao, “Conductance Titration of Single-Peptide Molecules,” J. Am. Chem. Soc. 2004, 126, 5370.] 14-1 Basic Concepts 273
1504T_ch14_270-297 1/23/06 11:35 Page 274 Power (watts) work per second P E I (IR ) I I 2 R Because q/s is the current, I, we can write P E I (14-8) A cell capable of delivering 1 ampere at a potential difference of 1 volt has a power output of 1 watt. Example Using Ohm’s Law + Battery 3.0 V – Electrons flow in this direction Resistor 100 Ω In the circuit in Figure 14-3, the battery generates a potential difference of 3.0 V, and the resistor has a resistance of 100 . We assume that the resistance of the wire connecting the battery and the resistor is negligible. How much current and how much power are delivered by the battery in this circuit? Figure 14-3 A circuit with a battery and a resistor. Benjamin Franklin investigated static electricity in the 1740s. 4 He thought electricity was a fluid that flows from a silk cloth to a glass rod when the rod is rubbed with the cloth. We now know that electrons flow from glass to silk. However, Franklin’s convention for the direction of electric current has been retained, so we say that current flows from positive to negative—in the opposite direction of electron flow. F is the Faraday constant (96 485 C/mol). E is electric potential difference measured in volts (V). E is the work (J) needed to move a coulomb of positive charge from one point to the other. n is moles of charge moved through the potential difference, E. I is electric current, measured in amperes (A). It is coulombs per second moving past a point in the circuit. R is resistance in ohms (). Units: A V/. Power is work per unit time (J/s) done by electricity moving through a circuit. Solution The current in this circuit is The power produced by the battery must be What happens to the power generated by the circuit? The energy appears as heat in the resistor. The power (90 mW) equals the rate at which heat is produced in the resistor. Here is a summary of symbols, units, and relations from the last few pages: Relation between charge and moles: Relation between work and voltage: Relation between free-energy difference and electric potential difference: Ohm’s law: Electric power: I E R 3.0 V 0.030 A 30 mA 100 P E I (3.0 V)(0.030 A) 90 mW q n F Charge Moles C/mole (coulombs, C) Work E q Joules, J Volts, V Coulombs G nFE Joules I E / R Current Volts Resistance (A) (V) (ohms, ) P work s E I Power J/s Volts Amperes (watts, W) A galvanic cell uses a spontaneous chemical reaction to generate electricity. 14-2 Galvanic Cells A galvanic cell (also called a voltaic cell) uses a spontaneous chemical reaction to generate electricity. To accomplish this, one reagent must be oxidized and another must be reduced. The two cannot be in contact, or electrons would flow directly from the reducing agent to the oxidizing agent. Instead, the oxidizing and reducing agents are physically separated, and electrons are forced to flow through an external circuit to go from one reactant to the other. Batteries 5 and fuel cells 6 are galvanic cells that consume their reactants to generate electricity. A battery has a static compartment filled with reactants. In a fuel cell, fresh reactants flow past the electrodes and products are continuously flushed from the cell. A Cell in Action Figure 14-4 shows a galvanic cell with two electrodes suspended in a solution of CdCl 2 . One electrode is cadmium; the other is metallic silver coated with solid AgCl. The reactions are Reduction: 2AgCl(s) 2e T 2Ag(s) 2Cl (aq) Oxidation: Cd(s) T Cd 2 (aq) 2e Net reaction: Cd(s) 2AgCl(s) T Cd 2 (aq) 2Ag(s) 2Cl (aq) 274 CHAPTER 14 Fundamentals of Electrochemistry
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