
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
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In the normal hydrogen electrode, the current flows through the electrode when the power difference
of the interface is +5 mV. Calculate the overvoltage n at pH = 2, if the equilibrium potential is -0.118 V.
In the normal hydrogen electrode, the balance potential difference in the interface is 0 and the current flow across the electrode when the interface potential difference is +5 mV. Explain briefly. Is the overvoltage 5 mV?
In the normal hydrogen electrode, the balance potential difference in the interface is 0 mV, the maximum potential is 5 mV. Explain briefly.
Chapter 5 Solutions
Chemistry: An Atoms-Focused Approach (Second Edition)
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- utron eutro cle TH tro (Na (b) Atoms are said to be electrically neutral. Explain. (c) Distinguish between the following: (i) Atomic number and mass number. (ii) Mass number and relative atomic mass. 2. An isotope Q, has 18 neutrons a mass number of 34. (a) (i) Draw the atomic structure of Q. (ii) Write its electron arrangement (b) To which period and group does Q belong? Explain your answer. (c) How does Q form its ion? Explain. 3. (a) Determine the relative atomic mass of the following elements = compositions occur in the proportions given. (i) Neon 20 21 22. Ne (90.92%), 10Ne (0.26%), and 10Ne (8.82%) (ii) Argon 36 38 40 18 Ar (0.34%), 18 Ar (0.06%) and 18 Ar (99.6%)arrow_forwardIn the normal hydrogen electrode, the balance potential difference in the interface is this, the maximum potential is 5 mV. Explain briefly.arrow_forwardThe 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?arrow_forward
- 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?arrow_forwardIn 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.arrow_forwardIn 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_forward
- In 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_forwardIf 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_forward
- Topic: 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_forwardWith the help of the Tafel line, it is estimated that the interchange density of the VO2+/VO2+ system on the carbon paper has a value of 3 mA cm-2. Calculate a) the current density if the voltage has a value of 1.6 mV and the temperature is 25°C. b) the beta value of the anódico process if the Tafel pendulum is 0.6 V at 25°C. Data: R = 8.314 JK-1mol-1, y F = 96485 C mol-1.arrow_forward
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