write the product favored reaction for the voltaic cellcontaining H 2 /H + and A

Question

write the product favored reaction for the voltaic cellcontaining H2/H+ and Ag/Ag+. a) and calculate the standard cell potential for the abovehydrogen/silver cell. b) then calculate the pH if the cell potential is 1.00V andall other components are at standard conditions. Use LeChatelier's Principle to explain the answer to b) write the product favored reaction for the voltaic cellcontaining H2/H+ and Ag/Ag+. a) and calculate the standard cell potential for the abovehydrogen/silver cell. b) then calculate the pH if the cell potential is 1.00V andall other components are at standard conditions. Use LeChatelier's Principle to explain the answer to b)

Explanation / Answer

Ag+   + e-     Ag            E_half = 0.7966 V H2 2H+   +2e-           E_half = 0 V -------------------------------- 2Ag+ + H2    Ag + 2H+ E0_cell = 0.7966 + 0 = 0.7966 V b) E = E0_cell - (R*T/nF)*ln{[H+]2/ [Ag+]2PH2]} n = 2 mol electron (# e- transferred in process); F = 96485 C/mol.R = 8.314 J/mol/K, T = 298K E = 1 V, E0_cell = 0.7966 V 1 = 0.7966 - (8.314*298/2/96485)*ln{[H+]2/[Ag+]2 PH2]} -15.8421624 = ln{[H+]2/ [Ag+]2PH2]} It is given that all other components are at standard conditions,so [Ag+] = 1, PH2 = 1 -15.8421624 = ln{[H+]2/1/1} = ln{[H+]2} [H+]2 = 1.3177599e-7 M [H+] = 0.000363009628 pH = -log[H+] = -log(0.000363009628) = 3.44008186 ~ 3.44 Le Chatelier: You reduced the concentration of [H+] from 1M (thestandard state) to 0.00036 M. H+ is a product, so you reduced theconcentration of a product. Notice that E = 1 V > E_0 cell (nowmore favorable for forward reaction), because you have reducedproduct concentration.

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