The material covered in this homework assignment applies to electrochemistry and

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_Ni^2+^3/a_Al^3-^2 = 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 of 8. What is the Eh of the system at 25 degree C? 1/8 NO_3^- + 5/4 H^+ + e^- doubleheadarrow 1/8 NH_4^+ + 3/8 H_2O Environmental issues arrising from arsenic toxicity are 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 boundary between the two ions on an Eh-pH diagram. There are several arsenic minerals, two of which are claudetite (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 are 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 of HS^- and SO_4^2. Thermodynamic Data Constants and Conversion Factors

Explanation / Answer

Ans 1: E= Eo- 0.0592/n x log K

EoCell for Al+3/Al = -1.67 & V EoCell for Ni+2/Ni= -0.257 V

Now, Al+3/Al is more electro negetive so its become oxiditation half cell so Al/Al+3 = 1.67

number of electrons change during reaction n=6

E= (1.67+(-0.257)) - 0.0592/6 x log 10-2

E = 1.413+0.019 = 1.432 V

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