1 (435)

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Chapter 19 Electrochemistry 435 = - = + + = [1(-394.4kJ) + - [1(-50.5kJ) + 2(0.0kJ)] = [-851.6kJ] = = 8.011 Also, because one C atom goes from an oxidation state of -4 to +4 and four atoms are going from 0 to -2, n = 8 and = -nF Rearrange to solve for Check: The units (V) are correct. The cell voltage is positive, which is consistent with a spontaneous reaction. 19.87 When iron corrodes or rusts, it oxidizes to Fe²⁺. For a metal to be able to protect iron, it must be more easily oxidized than iron or be below it in Table 19.1. (a) Zn and (c) Mn meet that criterion. Electrolytic Cells and Electrolysis 19.89 Given: electrolytic cell sketch Find: (a) Label the anode and cathode and indicate half-reactions, (b) indicate direction of electron flow, and (c) label battery terminals and calculate minimum voltage to drive reaction. Conceptual Plan: (a) Write two half-cell reactions and add electrons as needed to balance reactions. Look up half-reactions in Table 19.1. Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode: E°cell = - Choose the direction of the half-cell reactions so that