The material covered in this homework assignment applies to electrochemistry and
ID: 992345 • Letter: T
Question
The material covered in this homework assignment applies to electrochemistry and isotope geochemistry. Thermodynamic data and conversion factors that you may need are given. Show all your work. Combine the Al and Ni electrodes and calculate the EMF when a^3_Ni^2a/a^2_Al^3a = 10^-2. Consider the following half-cell redox reaction involving NO_3^- at a concentration of 10^-5 M (mol L^-1) and ammonium at a concentration of 10^-3 M in groundwater with a pH ol 8. What is the Eh of the system at 25 degree C? 1/8 NO_3^- + 5/4 H^+ + e^- leftrightarrow 1/8 NH_4^+ + 3/8 H_2O Environmental issues arrising from arsenic toxicity arc frequently associated with the oxidation of pyrite. which generates significant SO_4^2- and acid. Write a balanced half-cell reaction for the oxidate of HS^+ to SO_4^2-. Assuming aHS^+ = aSO_4^2- plot a phase boundrv between the two ions on an Eh-pH diagram. There are several arsenic minerals, two of which are claudetitc (As_2O_3) and orpiment (As_2S_3). Write a balance half-cell reaction between these two minerals, in an aqueous solution. Assume that the concentration of all aqueous species arc equal to 10^-6. Plot the phase boundary between the two minerals on your EH-pH diagram. And, comment on their stability fields with respect to the oxidation state HS^- and SO_4^2-. Thermodynamic data Constants and Conversion Factors 1 cal 4.184 J 1 bar 0.10 J cm^-3 a atm 1.01325 bar 1 cm^-3 0.10 J bar^-1 0 degree C 273.15 K R 8.3145 J mol^-1 K^-1 J 96485.309 JV^-1 mol^-1Explanation / Answer
Equation:
deltaG = deltaGº + RTlnQ
Q = a3Ni3+/a2Al3+ = 10-2
deltaGº = deltaGºAl3+ - deltaGNi2+
deltaGº = -485x103 J/mol - (-45.6x103 J/mol)
deltaGº = -439.4x103 J/mol
Calculate deltaG (for standard conditions):
deltaG = deltaGº + RTlnQ
deltaG = -439.4x103 J/mol + (8.3145 J/mol.K).(298 K)ln(10-2)
deltaG = -450.81x103 J/mol
Calculate E (emf) from deltaG:
deltaG =-nFE
E = -deltaG/nF
F, Faraday constant
n, transfered electrons
Determine n,
2Al(s) --> 2Al3+ + 6e-
3Ni2+ + 6e- --> 3Ni(s)
2Al(s) + 3Ni2+--> 2Al3+ + 3Ni(s)
n = 6
E = -(-450.81x103 J/mol )/(6).(96485.309 J/V.mol)
E = 0.779 V
EMF = 0.779 V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.