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In Tollen's test, glucose is oxidized to gluconic acid while Ag+ is reduced to Ag. Given the half-reactions: Ag+ + e- -> Ag, E°red = 0.80 V; C6H12O6 + H2O -> C6H12O7 + 2H+ + 2e-, oxidation half with E°red = -0.05 V for the reduction direction; and Ag(NH3)2+ + e- -> Ag(s) + 2NH3, E°red = -0.337 V. Using F/RT = 38.92 at 298 K, find ln K for the overall reaction 2Ag+ + C6H12O6 + H2O -> 2Ag(s) + C6H12O7 + 2H+.

1. 66.13
2. 58.38
3. 28.30
4. 46.29

Correct answer: 66.13

Solution

E°cell = E°cathode - E°anode = 0.80 - (-0.05) = 0.85 V. The number of electrons transferred is 2. Then ln K = n (F/RT) E°cell = 2 * 38.92 * 0.85 = 66.16, which rounds to about 66.13.

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