Please answer all the questions with all the steps for this multi-step Extra Cre
ID: 592619 • Letter: P
Question
Please answer all the questions with all the steps for this multi-step Extra Credit Question on Electro-Chemistry. Show all steps and provide explanations wherever possible.
a)A half-cell reads -0.5 V relative to an S.H.E. What will the potential be relative to a saturated Ag/AgCl electrode (.197v)?
b)What sign is the electrochemical potential of a spontaneous reaction?
c)Given .01 A for 32 sec, how many moles of Fe2+ can be reduced to Fe metal?
d)Given Eocathode = 0.1v, Eoanode = 0.2v, what is the Eocell?
e)Given this Eocell, what is Go for the reaction?
f)What happens at the equivalence point when Fe2+ is titrated with Ce4+?
g)Calculate the half-cell potential (E not E0) when 25 mL of 0.01 M Ce4+ is added to 15 mL of 0.01 M Fe2+ vs a S.H.E.
Ce4+ + e- = Ce3+ E0 = 1.70 V
Fe3+ + e- = Fe2+ E0 = 0.767 V
h)Draw the cell as it exists in the above question.
i)What is the purpose of a reference electrode in the pH electrode system?
j)What is the purpose of the salt bridge in a galvanic cell?
Explanation / Answer
(a) potential relative to Ag/AgCl=0.197-0.5=-0.303V
b) if the potential of the cell is positive, then it means that the reaction is spontaneous. Negative potential means that the reverse reaction is feasible.
(c) current I=0.01A, time t=32 seconds
Charge=0.01×32=0.32C
Number Of moles of electrons=0.32/96485=0.00033
One mole of Fe+2 is reduced to Fe by two moles of electrons
Number of moles of Fe+2 reduced=0.00033/2=0.000165
d) Eocell = Eocathode-Eoanode =0.1-0.2= -0.1V
e) Go=-nFEo where n=number of moles of elections, F= Faraday's constant, E=electrode potential. Assuming n=2 according to the information in the (c) part, we have G=-2×96485×(-0.1)=19.297KJ
(i) a reference electrode is used as a half cell to set up an electrochemical cell. It is stable and it's electrode potential is known. Example-standard hydrogen electrode (SHE)
(j) A salt bridge connects the two half cells in a electrochemical cell. It maintains the electrical neutrality of the cell. Electrons flow from anode to cathode. It completes the electrical circuit by connecting the two half cells and also prevents diffusion of solutions in the two half cells.