Consider the voltaic cell Ag(s)|Ag+( 0.012 M )||Fe3+( 0.058 M ),Fe2+( 0.044 M )|
ID: 964091 • Letter: C
Question
Consider the voltaic cell Ag(s)|Ag+( 0.012 M )||Fe3+( 0.058 M ),Fe2+( 0.044 M )|Pt(s)
Part A
What is Ecell initially?
Express your answer using three decimal places.
Part B (multiple choice)
As the cell operates, will increase, decrease, or remain constant with time?
Ecell remains constant
Part C (can't see)
Part D
What will be Ecell when [Ag+] has increased to 0.022 M ?
Express your answer using three decimal places.
Part E
What will be [Ag+] when Ecell= 0.015 V ?
Express your answer using two significant figures.
Part F
What are the ion concentrations when Ecell=0?
Express your answer using two significant figures. Enter your answers numerically separated by commas.
Ecell increases Ecell decreasesEcell remains constant
Part C (can't see)
Part D
What will be Ecell when [Ag+] has increased to 0.022 M ?
Express your answer using three decimal places.
Explanation / Answer
part A )
Eocell = 0.771 - 0.80 = - 0.029 M
overall reaction:
Ag + Fe+3 -----------------> Ag+ + Fe+2
Q = [Ag+][Fe+2]/[Fe+3]
Q = 0.012 x 0.044 / 0.058
Q = 9.1 x 10^-3
Ecell = Eo - 0.05916 / n * log Q
= -0.029 - 0.05916 * log (9.1 x 10^-3)
= 0.092 V
Ecell initially = 0.092 V
part B ) Ecell decreases
part D )
Q = [Ag+][Fe+2]/[Fe+3]
Q = 0.022 x 0.044 / 0.058
Q = 0.0167
Ecell = Eo - 0.05916 / n * log Q
= -0.029 - 0.05916 * log (0.0167)
= 0.076 V
Ecell = 0.076 V
part E)
Ecell = Eo - 0.05916 / n * log Q
0.015 = - 0.029 - 0.05916 / 1 * log Q
Q = 0.475
Q = [Ag+][Fe+2]/[Fe+3]
0.475 = [Ag+] x 0.044 / 0.058
[Ag+] = 0.63 M
part F )
each ion concentration = 1.0 M