Consider the voltaic cell, (Cr I Cr^3+ (0.200 M) II Pb^2+ (0.500 M) I Pb). E^deg
ID: 489482 • Letter: C
Question
Consider the voltaic cell, (Cr I Cr^3+ (0.200 M) II Pb^2+ (0.500 M) I Pb). E^degree _red (Cr^3_+/Cr) = -0.74 v and E^degree _red (Pb^2+/Pb) = -0.126 v Calculate Delta E_cell and Delta E^degree _cell for this voltaic cell. As the redox reaction proceeds the Pb^2+ is slowly consumed. When the [Pb^2+] reaches 0.0500 M, what will be the values of [Cr^3+] and Delta E_cell? Assume both half-cells started with 100 mL of solution. A 0.5305-g sample of an unknown alkali metal chloride salt was dissolved in water and an electric current was passed through the solution until no more chlorine gas was evolved. The chlorine gas was dried and then transferred to a 114.2 mL container at 25^degree C producing a pressure of 739.0 torr. Assuming all the chloride anions were converted to chlorine gas, identify the alkali metal and give the formula of the metal chloride salt. If the current used for this electrolysis process was 0.500 amperes, then calculate the amount of time required to oxidize all of the chloride ions in the original salt solution.Explanation / Answer
Q1.
E°Cell = Ered - Eox
clearly, Cr+3 wll oxidize and Pb+2 will reduce due to their potential
so
E°cell = (-0.126 - -0.74) = 0.614 V
so
Ecell = E°cell - 0.0592/(n) * log(Q)
Q = [poductS]/[Reactans]
Cr(s) --> Cr+3 + 3e-
Pb+2 + 2e- --> Pb(s)
balance
2Cr(s) --> 2Cr+3 + 6e-
3Pb+2 + 6e- --> 3Pb(s)
Q = [Cr+3]^2/ [Pb+2]^3
Q = (0.2^2) / (0.5^3) = 0.32
n = 6 electrons
Ecell = E°cell - 0.0592/(n) * log(Q)
Ecell =0.614 - 0.0592/(6) * log(0.32)
Ecell = 0.61888V
when Pb+2 = 0.05 find Cr+3 and Ecell:
Ecell = E°cell - 0.0592/(n) * log(Q)
Ecell = 0.614 - 0.0592/(6) * log([Cr+3]^2/ [Pb+2]^3
[Cr+3] = 0 + 3x
[Pb+2] = 0.5 - 2x = 0.05
x = (0.05-0.5)/(-2) = 0.225
[Cr+3] = 0.2 + 3x = 0.2 + 3*0.225 = 0.875
Ecell = 0.614 - 0.0592/(6) * log([Cr+3]^2/ [Pb+2]^3
Ecell = 0.614 - 0.0592/(6) * log((0.875^2)/(0.05^3)) = 0.576 V
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