a) If this electrode is the rigth-hand electrode and the SHE is the left-hand el
ID: 853674 • Letter: A
Question
a) If this electrode is the rigth-hand electrode and the SHE is the left-hand electrode, the cell potential is -0.245 V. What is the activity of the sulfate ion in this cell?
b) Calculate the mean ionic activity of the sulfuric acid in the cell
Pt (s) | H2 (g, 1 atm) | H2SO4 (aq, a) | PbSO4 (s) | Pb (s)
if the cell potential is -0.220 V. (Note:the left-hand electrode is not the SHE)
For the electrode SO4^2 - (aq, aSO4^2 - ) | PbSO4 (s) | Pb (S), phi= - 0.356 V a) If this electrode is the rigth - hand electrode and the SHE is the left - hand electrode, the cell potential is - 0.245 V. What is the activity of the sulfate ion in this cell? b) Calculate the mean ionic activity of the sulfuric acid in the cell Pt (s) | H2 (g, 1 atm) | H2SO4 (aq, a) | PbSO4 (s) | Pb (s) if the cell potential is - 0.220 V. (Note:the left - hand electrode is not the SHE)Explanation / Answer
From the cell notation the reaction takes place at catode is
PbSO4 +2e- -----------------------> Pb +SO42-
We know
Ecell= E0 - (0.0591/n)logK at 25C ( activity = concentration for very dilute solutions)
-0.245= -0.356-(0.0591/2) log([aPb2+][aSO42-] /[aPbSO4]) ( activity = concentration for very
3.756=-log[aSO42-] [aPb2+]=1 and [aPbSO4]=1
Activity of the sulphate ion [aSO42-]= 1.753*10^-4
2.) Apply same above mentioned methode by chaning cell voltage
-0.22=0.356-0.02955 log[aSO42-]
finally [aSO42-]=[SO42-]=2.5*10^-5M
So H2SO4 concentration = 2.5*10^-5M
Mean activity a
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.