a) For a reaction 2H+ + 2e- + 0.5O2 = H2O. Standard E = 1.23. If the electrode p

Question

a) For a reaction 2H+ + 2e- + 0.5O2 = H2O. Standard E = 1.23. If the electrode potential is reduced to 1.00V, will this bias the reaction in the forward or reverse direction?

b) For the reaction 2H+ + 2e- = H2. E = 0.00V. if the potential is increased to 0.01V, will this bias the reaction in the forward or reverse direction?

c) If the electrode in (a) is connected to the electrode in (b), what is the cell potential when both of the electrodes is at equlibrium?

d) Consider your answer for (a) and (b), what is the cell potential when both of the electrodes is shifted from equilibrium?

e) Consider your answer for part (d), what is the total potential loss in that case. Support your answers for all the question parts with equations and I-V diagrams.

p.s Please show your work and explanation, as I'd love to know what I'm doing and not just copy the answers

Explanation / Answer

a)As resultant E value=1.23-1.00 is greater than 0, the reaction is in foward direction.

b)As resultant E=0.01V, Reaction moves forward

c)E at equilibrium is always 0.

d) E(net)=E(Cathode)-E(anode)=0.23-0.01=0.22V.

e)Total potential loss is same is total voltage initially=0.22V.

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