(a) For reactions (b) and (c) in Problem 4, convert the given E° values (i.e., t

Question

(a) For reactions (b) and (c) in Problem 4, convert the given E° values (i.e., the standard potential) to pe° (assume constant temperature).

H2S(aq) + 4H2O = SO4 2- + 10H+ + 8e- E° = +0.76 v

(b) Now write an equation for a straight line (y = mx + b) using the Nernst expression for pe (rather than Eh), with x = pH, y = pe, and b = the remaining terms for each reaction (you don't have to solve the equation, just write it as the equation for a straight line). How does the form of the Nernst equation differ when written as pe? What is the difference in the scale of the y-axis for Eh versus pe?

Explanation / Answer

E = E* + 0.0591/n log [H+][SO42-]/[H2S]

This is the nerst equation and in this case as we want a straight line curve (y=mx+b)

We know that log conc will be a constant value and the varying thing is the pH which we will remove from the log terms

The equation will be

Eh = E*h + 0.0591/n log Concentartions - 0.0591/n * 10 pH

so from the above equation we can surely see that it come out to be a straight line with pH and standard potential varying

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