
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN: 9781938168390
Author: Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher: OpenStax
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Chapter 6, Problem 42E
State the Heisenberg uncertainty principle. Describe briefly what the principle implies.
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The electrode balance potential is -0.118 V and the interface potential difference
is +5 mV. The overvoltage n will be 0.005 - (-0.118) = 0.123 V. Is it correct?
In the electrode Pt, H2(1 atm) | H+(a=1), if the electrode balance potential is -0.118 V and the interface potential difference is +5 mV. The current voltage will be 0.005 - (-0.118) = 0.123 V ¿Correcto?
In the electrode Pt, H2(1 atm) | H+(a=1) at 298K is 0.79 mA cm-2. If the balance potential of the electrode is -0.118 V and the potential difference of the interface is +5 mV. Determine its potential.
Chapter 6 Solutions
Chemistry by OpenStax (2015-05-04)
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- In one electrode: Pt, H2(1 atm) | H+(a=1), the interchange current density at 298K is 0.79 mA·cm-2. If the voltage difference of the interface is +5 mV. What will be the correct intensity at pH = 2?. Maximum transfer voltage and beta = 0.5.arrow_forwardIn a Pt electrode, H2(1 atm) | H+(a=1), the interchange current density of an electrode is 0.79 mA cm-2. ¿Qué corriente flow across the electrode of área 5 cm2 when the difference in potential of the interface is +5 mV?.arrow_forwardIf the current voltage is n = 0.14 V, indicate which of the 2 voltage formulas of the ley of Tafel must be applied i a a) == exp (1-B). xp[(1 - ß³): Fn Fn a b) == exp B RT RTarrow_forward
- If the current voltage is n = 0.14 V. Indicate which of the 2 formulas must be applied a) = a T = i exp[(1 - p) F Fn Fn b) i==exp B RTarrow_forwardTopic: Photochemistry and Photophysics of Supramoleculesarrow_forwardTwo cations that exchange an electron in an interface, the exchange density is worth 1.39 mA/cm2 and the current density is worth 15 mA/cm2 at 25°C. If the overvoltage is 0.14 V, calculate the reaction rate and symmetry factor. Data: R = 8,314 J mol-1 k-1: F = 96500 Carrow_forward
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