A sodium ion-selective electrode increased in voltage by 59.16 mV when the sodium cation concentration was increased by a factor of 10. A S2-ion-selective electrode measured a decrease of 29.58 mV when the sulfide concentration increased by a factor of 10. Why? Select all that apply A. Cations always increase the measured potential by z x 59.16 mV where z is the charge on the cation. UB. Anions always decrease the measured potential by z x 59.16 mV where z is the charge on the anion. C. Cations always increase the measured potential by 59.16 mV/z, where z is the charge on the cation. D. Anions always decrease the measured potential by 59.16 mV/z, where z is the charge on the anion.

A sodium ion-selective electrode increased in voltage by 59.16 mV when the sodium cation concentration was increased by a factor of 10. A S2-ion-selective electrode measured a decrease of 29.58 mV when the sulfide concentration increased by a factor of 10. Why? Select all that apply A. Cations always increase the measured potential by z x 59.16 mV where z is the charge on the cation. UB. Anions always decrease the measured potential by z x 59.16 mV where z is the charge on the anion. C. Cations always increase the measured potential by 59.16 mV/z, where z is the charge on the cation. D. Anions always decrease the measured potential by 59.16 mV/z, where z is the charge on the anion.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

2. You are given an electrochemical cell where the chemical reaction involves a silver complex, and a silver metal bar is provided as the positive electrode. The electrochemical potential for this cell is given by a. b. E=Eº 1 [Ag] -0.0592 (volts).log- Ag /Ag For a test solution with a silver concentration of 0.001 M, the electrochemical cell potential is 0.20 volts. Find the silver electrode potential EºAg-Ag. The same cell from (a) will be used to measure the electrochemical potential for a new silver complex in equilibrium. If the measured cell potential is 0.324 volts, what is the silver concentration for this new silver complex at a chemical equilibrium state?
Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K.
3. Tanjiro, a Chem 17.1 student, found a prehistoric katana during his trip to Japan. He inspected the katana and realized that it was made out of gold. Upon closer inspection, Tanjiro noticed that it had a lot of chips and scratches, signifying that it had been used in battle long ago. Wanting to restore the katana to its original state, Tanjiro used his knowledge in Chem 17.1 to electroplate gold into the surface of the katana using an ore his father found on their backyard. Tanjiro proceeded with the electroplating process, however, he realized that the concentration of the gold ions present in the solution is unknown, making the electroplating process difficult to plan. He then decided to use a galvanic cell to obtain the gold ion concentration. He used the half-cell reactions below: Au³+ (aq) + 3e → Au(s) E° = +1.520V E° = -0.230V Ni²+ (aq) + 2e → Ni(s) Given: [Ni²+] = 0.050 M FMNI = 58.69 g/mol FMAU = 196.97 g/mol
I need the answer as soon as possible
Calculate the cell potential for a cell at 29°C with a Cu2+ concentration of 0.017 M and a Zn²+ concentration of 0.068 M. Zn(s) + Cu2*(aq) =Zn2*(aq) + Cu(s) E°cell = +1.10 V Round your answer to 3 decimal places.
Calculate the cell potential for a cell at 31°C with a Cu2+ concentration of 0.011 M and a Zn2+ concentration of 0.056 M. Zn(s) + Cu2*(aq) = Zn2*(aq) + Cu(s) E°cell = +1.10 V Round your answer to 3 decimal places.
When a fluoride selective electrode was immersed in a solution that is 1.0 x 10-4 M in F-, the potential that developed (versus SCE) was 41.5 mV. What will be the potential that will develop if the same electrode is immersed in a solution that is 5.0 x 10-5 M in fluoride? -51 mV -47 mV -59 mV -32 mV
The reduction potential to the metal of Ba 2+ is -1.57 V and that of Cu 2+ is +0.34 V. The cell potential for the reaction Ba(s) + Cu 2+ (aq) → Ba 2+ (aq) + Cu(s) is therefore: a. (+1.57) + (+0.34) b. (+1.57) + (-0.34) c. (-1.57) + (+0.34) d. 2 x (-1.57) + 2 x (0.34) e. (-1.57) + (0.34)
Calculate the cell potential for a cell at 40oC with a Cu2+ concentration of 0.014 M and a Zn2+ concentration of 0.069 M. Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s) Eocell = +1.10 V Round your answer to 3 decimal places.
Calculate the cell potential for a cell at 43oC with a Cu2+ concentration of 0.012 M and a Zn2+ concentration of 0.063 M. Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s) Eocell = +1.10 V Round your answer to 3 decimal places.
What is the emf of a galvanic cell consisting of a 4/=7/4/ half-cell and a e#/57/5= halfcell if [4/=7] = 0.20 t,[57] = 0.16 t , and e § = 0.80 atm? (Masterton et al., 201
Find the potentials of the following electrochemical cell:Cd | Cd2+, M = 0.10 ‖ Ni2+, M = 0.50 | Ni