As researchers look for alternative fuels, the production of H_2 is being extens
ID: 509846 • Letter: A
Question
As researchers look for alternative fuels, the production of H_2 is being extensively studied. The following example shows a reduction reaction that can be used to produce H_2 under basic conditions: 2H_2O(l) + 2e^- rightarrow H_2(g) + 2OH^-(aq) E degree (g) = -0.83 V a) What is the Nernst expression for this half-cell? (b) Write the Nernst expression so that the influence of the [OH^-] appears as a function of pH. (assume activity of H_2 = 4.0 times 10^-5) (c) What is the value of E for this reaction when the pH of the solution is maintained at 7.00? (d) What is the value of E for this reaction when the pH of the solution is maintained at 10.00? (e) What trend is noticed for this reaction as the pH is increased? Using LeChatelier's Principle explain the basis of this trend. (f) What is the effect on E on this reaction when the pH is lowered by one pH unit?Explanation / Answer
2H2O (l) + 2e- ------> H2(g) + 2OH-
(a) nernst equation is :
E= E^o - 0.059/2 log [a(H2)* {a(OH-)}^2]
(b) [OH-] = Kw/[H+]
E= E^o - 0.059/2 log [a(H2)* {a(OH-)}^2]
= -0.83 - 0.059/2 log (4*10^-5 * {Kw/[H+]}^2)
= -0.83 - 0.059/2 log ( 4*10^-33/[H+]^2)
=- 0.83 + 0.96 - log(1/[H+]^2)
= 0.13 + 2log[H+]
= 0.13- 2pH
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(c) when the value of pH is 7 , E= -13.87 V
(d) at pH = 10, E = -19.87 V
(e) at higher pH the reaction becomes difficult. This can be justified from Le Chatelier principle as increasing OH- will shift the equilibrium to left hand side.
(g) Lower the pH will make the E less negative and equilibrium will shift to right
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