Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum. Copyright © Cengage Learning. All rights reserved

Maximum Cell Potential Directly related to the free energy difference between the reactants and the products in the cell. ΔG° = –nFE° F = 96,485 C/mol e– Copyright © Cengage Learning. All rights reserved

A Concentration Cell Copyright © Cengage Learning. All rights reserved

Nernst Equation The relationship between cell potential and concentrations of cell components At 25°C: or (at equilibrium) Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Explain the difference between E and E°. When is E equal to zero? When is E° equal to zero? ε is the cell potential at any condition, and ε is the cell potential under standard conditions (1.0 M or 1 atm, and 25C). ε equals zero when the cell is in equilibrium ("dead" battery). ε is equal to zero for a concentration cell. Copyright © Cengage Learning. All rights reserved

EXERCISE! A concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and 1.00 × 10-4 M in the two half-cells. Calculate the potential of this cell at 25°C. The cell potential equals 0.118 V. ε = ε° - (0.0591/n)logQ. For a concentration cell, ε° = 0. ε = 0 - (0.0591/2)log(1.0×10-4 / 1.0) = 0.118 V Copyright © Cengage Learning. All rights reserved

You make a galvanic cell at 25°C containing: CONCEPT CHECK! You make a galvanic cell at 25°C containing: A nickel electrode in 1.0 M Ni2+(aq) A silver electrode in 1.0 M Ag+(aq) Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential. Nickel is the anode, silver is the cathode (electrons flow from nickel to silver). To find the cell potential, use the Nernst equation: ε = ε° - (0.0591/n)logQ. The overall equation is 2Ag+ + Ni → 2Ag + Ni2+. Therefore , Q = [Ni2+] / [Ag+]2 = (1.0 M) / (1.0 M)2. ε = 1.03 V – (0.0591/2)log(1) = 1.03 V. Copyright © Cengage Learning. All rights reserved

Forcing a current through a cell to produce a chemical change for which the cell potential is negative. Copyright © Cengage Learning. All rights reserved

Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time  quantity of charge  moles of electrons  moles of analyte  grams of analyte Copyright © Cengage Learning. All rights reserved

Stoichiometry of Electrolysis current and time  quantity of charge Coulombs of charge = amps (C/s) × seconds (s) quantity of charge  moles of electrons Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO3)3. What is the metal? The metal has a molar mass of 197 g/mol. The metal is gold (Au). To find the molar mass, divide the number of grams of the metal by the number of moles of metal. To find the moles of metal: 52.8 sec × (2.00 C/sec) × (1 mol e–/96485 C) × (1 mol M/3 mole e–) = 3.65 × 10–4 mol M To find the molar mass: 0.0719 g / 3.65 × 10–4 mol M = 197 g/mol Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Do the metals form on the cathode or the anode? Explain. Use the reduction potentials from Table 18.1. Once that voltage has been surpassed, the metal will plate out. They plate out on the cathode (since they are reduced). Order: Cu2+, Pb2+, Sn2+, Ni2+, Zn2+ Copyright © Cengage Learning. All rights reserved