Chlorine dioxide ClO 2 is produced by the reaction 2NaClO 2 aq Cl 2 g 2ClO 2 g

# Chlorine dioxide clo 2 is produced by the reaction

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Chlorine dioxide (ClO 2 ) is produced by the reaction 2NaClO 2 (aq) + Cl 2 (g) → 2ClO 2 (g) + 2NaCl(aq) Using data from Table 17.1, calculate E o and ΔG o at 25 o C for the production of ClO 2 . 2ClO 2 - (aq) + Cl 2 (g) → 2ClO 2 (g) + 2Cl - (aq) Table 17.1 E o ClO 2 + e - → ClO 2 - +0.954 Cl 2 + 2e - → 2Cl - +1.36 2(ClO 2 - → ClO 2 + 1e - ) -0.954 Cl 2 + 2e - → 2Cl - +1.36 2ClO 2 - + Cl 2 → 2ClO 2 + 2Cl - +0.406 V = E o cell E o cell = +0.406 volts ΔG o = -nFE o cell = -2 x 96485 x 0.406 = -7.83 x 10 4 J Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 53
H&P Example 18.8 Calculate the values of K eq for the reaction Cu(s) + 2Ag + (1M) Cu 2+ (1M) + 2Ag(s) Cu 2+ + 2e - → Cu E o = +0.340 Ag + +1e - → Ag E o = +0.800 Cu → Cu 2+ + 2e - E o = -0.340 Ag + +1e - → Ag E o = +0.800 Cu → Cu 2+ + 2e - E o = -0.340 2(Ag + +1e - → Ag) E o = +0.800 Cu + 2Ag + Cu 2+ + 2Ag E o cell = +0.460V E o cell = (0.0592/n)log(K) 0.460 = (0.05915/2)log(K) K = 3.58 x 10 15 = 4 x 10 15 Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 54
ET: Discuss extensive vs. intensive property. An intensive property is a physical property of a system that does not depend on the system size or the amount of material in the system. Standard potential voltage is intensive because it is Joules/Coulomb, which is energy per electron. Note that when a half-cell reaction is doubled, the substance doubles (which contains energy therefore the Joules doubles), but also the electrons transferred doubles. Since volts is Joules/Coulomb, and they both double, then the volts don’t change. On the other hand, volts is an extensive property when just the Joules are increased, e.g., in Nernst’s equation going from standard conditions (1M) to non-standard conditions (e.g., 2M). Note that here the electrons transferred don’t change, while the concentration of reactant increases. E cell = E o cell -(0.0592/2)log(1/1). Double the concentration of product provides: E cell = E o cell -(0.0592/2)log(2/1), where the product has increased but the number of electrons transferred hasn’t changed. Z&Z 51 NERNST EQUATION Given new concentrations and E o , find qualitative effect on E o . Given half-cell standard voltages, find E o cell; then given non-standard concentrations, qualitatively and quantitatively find E A voltaic cell is based on the following half-reactions at 25 o C and standard cond. Predict whether E cell is larger or smaller than E o cell for the following: [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M, [Ag + ] = 1.0 M Standard state reduction potentials E o (V) H 2 O 2 + 2H + + 2e - 2H 2 O 1.78 Ag + + e - Ag 0.80 Write spontaneous equation under standard state conditions . E o (V) H 2 O 2 (1M) + 2H + (1M) + 2e - 2H 2 O( l ) 1.78 2(Ag(s)   Ag + (1M) + e - ) -0.80 2Ag(s) + H 2 O 2 (1.0M) + 2H + (1.0M) 2Ag + (1.0M) + 2H 2 O( l ) +0.98V Actual conditions H 2 O 2 (2.0M) + 2H + (2.0M) + 2e - 2H 2 O( l ) ? 2(Ag(s)   Ag + (1M) + e - ) -0.80 2Ag(s) + H 2 O 2 (2.0M) + 2H + (2.0M) 2Ag + (1.0M) + 2H 2 O( l ) ?V Qualitative : Based on the high concentration of reactants, vs. products, according to the LeChatelier principle the reaction will go further to the right, i.e., be more spontaneous than (i.e., greater than) +0.98V.

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