The cell potential of a redox reaction occurring in an electrochemical cell unde
ID: 526997 • Letter: T
Question
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration conditions can be determined from the standard cell potential of the cell using the Nernst equation
E = E°
ln Q
where E is the cell potential of the cell, E° is the standard cell potential of the cell, R is the gas constant, T is the temperature in kelvin, n is the moles of electrons transferred in the reaction, and Q is the reaction quotient. Use this relationship to answer the problem below.
For the following oxidation-reduction reaction 2 Au3+(aq) + 3 Mn(s) 2 Au(s) + 3 Mn2+(aq) the standard cell potential is
2.68 V.
What is the actual cell potential of the cell if the temperature is 311 K, the initial Au3+ concentration is 0.00121 M, and the initial Mn2+ concentration is 0.0349 M? (Note that the reaction involves the transfer of 6 moles of electrons, and the reaction quotient is 29.0.)
Identify each of the known and unknown variables and constants in the equation. (Give your answer to the full known precision.)
The value for E° is given.
E° = V
R is the constant.
R = J/mol·K
The value for T is given.
T = K
F is Faraday's constant.
F = C/mol
The value for n is given.
n =
The value for Q is given.
Q =
B) Perform any necessary unit conversions.
The units of Faraday's constant must be converted to J/V·mol so that the joules will cancel when substituted into the equation. (Enter an unrounded value.)
9.64853415104 C/mol 1 J/C/1V = J/V·mol
Explanation / Answer
Ecell = E°cell - (RT/nF)ln(Q)
E°cell = 2.68 VR=8.3145 (J/mol)(K) T= 38C or 311K n= 6 (as two e- ) F= 96,485 c/mol = Q= 29
Ecell = 2.68 V – (8.3145 J/mol.K*311 K)/(6 mole*96485 c/mole)*ln(29)
= 2.68-((2585.8095)/578910))*3.3672 = 2.68 – 0.0150
= 2.665 V
B)
Let us convert units of faraday’s constant to J/V.mol
F’=9.64853415*104C/mol*1 J/C/1V =
Ecell = 2.68 V – (8.3145 J/mol.K*311 K)/(6 mole*1003.4475516 J/V.mole)*ln(29)
= 2.68 V – ((2585.8095 J/mol)/6020.6853096 J/V) * 3.3672
= 2.68 V – (0.42948756)*3.3672
= 2.68 V – 1.4461705
Ecell = 1.2338295
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