412 tential of a Fe electrode in a solution of 0.05 M FeCh pH 2.0, E0-0.036 V at 25 °C.
A galvanic cell is built at 25 ºC using a Cr electrode inside a 2.0 M Cr3+ solution (100 mL) and a Ni electrode inside a 0.5 M Ni2+ solution (100 mL). Both solutions are connected with a salt bridge. a) (0.5 p) Write the cell notation and the global redox reaction. Indicate the anode and the cathode of the cell. Determine the cell potential with the given concentrations. b) (0.5 p) Determine how many of the previous cells must...
Given the following electrode potentials at 25°C Fe3+ + e- → Fe2+ E° = 0.171 V Fe2+ + 2e-→ Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- → Fe(s) Select one: a.-0.236 b. 0.081 c.-0.036 d. -0.211 e. 0.121
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- Fe(s) Fe2+ + Select one: a. -0.132 b. -0.036 c. 0.081 d.-0.211 e. 0.103
A solution of 0.2 M dehydroascorbate and 0.2 M ascorbate (E0+ = +0.06 V) was mixed with an equal volume of a solution containing 0.01 M acetaldehyde and 0.01 M ethanol (E 0 = -0.163 V) at 25° C and pH 7. Calculate the AE for the possible reaction.
An uncalibrated pH electrode produces a voltage of 0.184 V for a solution of pH 5.89 at 25 degrees Celcius. If the same electrode is used under the same conditions to measure a solution of unknown pH and 0.376 V is measured, what is the pH of the second solution?
1. Calculate the pH and pOH of a 0.036 M solution of HBr.
(a) When the difference in pH across the membrane of a glass electrode at 25 °C is 4.63 pH units, how much voltage is generated by the pH gradient? (b) What would be the voltage for the same pH difference at 37 °C? (c) The pH electrode has a selectivity coefficient k for Na+ of 10^-8 . What is the measured the pH value when the solution pH is at 9.0 with a Na+ concentration of 0.1 M?
7.53 Given the following electrode potentials at 25°C, Fe3+ +e- = Fe2+ E' = 0.771 V Fe2+ + e = Fe(s) E = -0.440 V calculate the electrode potential for [Fe3+ + e = {Fe E; = ? for
Calculate the approximate pH of an aqueous solution of H2S , 0.05 M, and H3PO4, 0.21 M at 25°C. (Note: consider one decimal place for the pH answer.) pKa H2S/HS- = 7.02, pKa HS-/S2- = 13.99 pKaH3PO4/ H2PO4- =2.15, pKaH2PO4-/HPO42-=7.15, pKa HPO42-/ PO43- =12.38
The solubility product of Fe(OH)3 at 25°C is 10-38.1 and . Calculate the standard emf (E”) for OH", Fe(OH)31Fe electrode. Given: Fe(OH)3 - Fe+3+3 OH Fe+3+3 € → Fe (Fe3+1Fe) = -0.036 V Fe(OH)3 + 3 e Fe(OH", Fe(OH)3|Fe) E. OH-, Fe(OH)3 Fe Ksp = 10-38.1 E festlife 0.642 V 0.785 V -0.785 V -0.642 V