Use standard reduction potentials to calculate the equilibrium constant for the reaction: Hg2+(aq) + 2Cr2+(aq)→ Hg(l) + 2Cr3+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: ______ G° for this reaction would be _________ (greater or less )  than zero.

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Use standard reduction potentials to calculate the equilibrium constant for the reaction:
Hg2+(aq) + 2Cr2+(aq)→ Hg(l) + 2Cr3+(aq)


Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.

Equilibrium constant: ______

G° for this reaction would be _________ (greater or less )  than zero.

Relationship between E°cell and K
The equilibrium constant for a cell reaction can be calculated from the standard cell voltage. The relationship between them is:
RT
E°.
cell
In K
nF
where
standard cell potential
= ideal gas constant
kelvin temperature
= number of moles of electrons for the reaction as written
F = charge carried by 1 mol of electrons
K = equilibrium constant
E°
R
T
n
It is customary to use the equation in a form where numerical values are substituted for R, T and F at a temperature of 25 °C.
For
R =
8.314 J mol-'K!
T
298.15 K
F =
96,485 J V' mol"1
(8.314 J mol K-)(298.15 K)
RT
= 0.0257 V
F
96,485 J V-l mol!
and the equation with the potential in volts is:
0.0257
Previous
Next
E° cell
In K
natural logarithm
Transcribed Image Text:Relationship between E°cell and K The equilibrium constant for a cell reaction can be calculated from the standard cell voltage. The relationship between them is: RT E°. cell In K nF where standard cell potential = ideal gas constant kelvin temperature = number of moles of electrons for the reaction as written F = charge carried by 1 mol of electrons K = equilibrium constant E° R T n It is customary to use the equation in a form where numerical values are substituted for R, T and F at a temperature of 25 °C. For R = 8.314 J mol-'K! T 298.15 K F = 96,485 J V' mol"1 (8.314 J mol K-)(298.15 K) RT = 0.0257 V F 96,485 J V-l mol! and the equation with the potential in volts is: 0.0257 Previous Next E° cell In K natural logarithm
R =
8.314 J mol"'K!
T
298.15 K
F =
96,485 J V-' mol"1
(8.314 J mol K-)(298.15 K)
RT
0.0257 V
96,485 J V' mol
F
and the equation with the potential in volts is:
0.0257
E°
cell
In K
natural logarithm
Sometimes the base-10 logarithm is used and the substitution of 2.303 log for In is made. (2.303 x 0.0257 = 0.0592) Then, the equation
for base-10 logs at 25 °C is:
0.0592
E° cell
log K
base-10 logarithm
n
A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm.
Transcribed Image Text:R = 8.314 J mol"'K! T 298.15 K F = 96,485 J V-' mol"1 (8.314 J mol K-)(298.15 K) RT 0.0257 V 96,485 J V' mol F and the equation with the potential in volts is: 0.0257 E° cell In K natural logarithm Sometimes the base-10 logarithm is used and the substitution of 2.303 log for In is made. (2.303 x 0.0257 = 0.0592) Then, the equation for base-10 logs at 25 °C is: 0.0592 E° cell log K base-10 logarithm n A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm.
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