Predict the standard cell potential and calculate the standard reaction Gibbs free energy for galvanic cells having the following cell reactions:
a.)3Zn + 2Bi3+ = 3Zn2+ + 2Bi
b.)2H2 + O2 = 2H20 in acidic solution
c.)2H2 + O2 = 2H20 in basic solution
d.)3Au+ = 2Au + Au3+
I tried all of them and I could not figure any of them out. I am so confused. Thanks in advance.
You will find Zn^2+ + 2e ==> Zn in the standard reduction tables. You want the reverse of this so simply change the sign.
Then look up Bi^3+ + 3e ==> Bi
Zn number (with changed sign) + Bi number = Eocell.
Then dG = -nEF
n in this case is 6, E is from above, F is 96,485 coulombs.
To predict the standard cell potential and calculate the standard reaction Gibbs free energy, we will use the Nernst equation and standard reduction potentials. The standard reduction potentials are given in volts (V) at 25°C.
a.) The balanced cell reaction is: 3Zn + 2Bi3+ → 3Zn2+ + 2Bi
The half-reactions are:
Zn2+ + 2e- → Zn (reduction)
Bi3+ + 3e- → Bi (oxidation)
The standard reduction potentials for these reactions are:
Zn2+/Zn: -0.76 V
Bi3+/Bi: -0.82 V
The standard cell potential (E°cell) can be calculated by subtracting the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction:
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = -0.76 V - (-0.82 V)
E°cell = 0.06 V
The standard reaction Gibbs free energy (ΔG°) can be calculated using the equation:
ΔG° = -nF E°cell
where n is the number of moles of electrons transferred and F is Faraday's constant (96,485 C/mol).
In this case, n = 6 moles of electrons transferred (3 from Zn oxidation and 2 from Bi reduction):
ΔG° = -6(96485 C/mol)(0.06 V)
ΔG° = -34864.2 J/mol
b.) The balanced cell reaction is: 2H2 + O2 → 2H2O
The half-reactions are:
H+ + e- → 1/2H2 (reduction)
O2 + 4H+ + 4e- → 2H2O (oxidation)
The standard reduction potentials for these reactions are:
1/2H2/H+: 0 V
O2/H2O: 1.23 V
To get the complete reaction, we need to balance the half-reactions with respect to electrons:
2(H+ + e- → 1/2H2) + O2 + 4H+ + 4e- → 2H2O
2H+ + O2 → H2O
The standard cell potential can be calculated in the same way as in part a):
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = 0 V - 1.23 V
E°cell = -1.23 V
The number of moles of electrons transferred is 4 (from the reduction half-reaction):
ΔG° = -4(96485 C/mol)(-1.23 V)
ΔG° = 474271.4 J/mol
c.) The balanced cell reaction is the same as in part b):
2H2 + O2 → 2H2O
The standard cell potential and ΔG° will also be the same as in part b):
E°cell = -1.23 V
ΔG° = 474271.4 J/mol
d.) The balanced cell reaction is: 3Au+ → 2Au + Au3+
The half-reactions are:
Au3+ + 3e- → Au (reduction)
Au+ → Au+3 + e- (oxidation)
The standard reduction potentials for these reactions are:
Au+/Au: +1.98 V
Au3+/Au: +1.40 V
To get the complete reaction, we need to balance the half-reactions with respect to electrons:
3(Au3+ + 3e- → Au) + 2(Au+ → Au3+ + e-) → 2Au + Au3+
3Au3+ + 3Au+ → 5Au + 3Au3+
The standard cell potential can be calculated:
E°cell = E°red(reduction) - E°red(oxidation)
E°cell = +1.98 V - (+1.40 V)
E°cell = +0.58 V
The number of moles of electrons transferred is 3 (from the oxidation half-reaction):
ΔG° = -3(96485 C/mol)(0.58 V)
ΔG° = -168062.7 J/mol
I hope this helps! Let me know if you have any further questions.
To predict the standard cell potential and calculate the standard reaction Gibbs free energy for galvanic cells, you need to use the Nernst equation and the standard reduction potentials.
The Nernst equation relates the cell potential to the concentrations of the species involved in the redox reaction. It is given as:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell = cell potential (in volts)
E°cell = standard cell potential (in volts)
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons transferred in the balanced redox equation
F = Faraday constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient
The reaction quotient (Q) is the ratio of concentrations of products to reactants, raised to the power of their stoichiometric coefficients. In a galvanic cell, it is the ratio of product concentrations to reactant concentrations.
To calculate the standard reaction Gibbs free energy (ΔG°), you can use the equation:
ΔG° = -nF E°cell
Now, let's apply this information to each of the given cell reactions:
a) 3Zn + 2Bi3+ --> 3Zn2+ + 2Bi
To determine the standard cell potential (E°cell), we need to know the standard reduction potentials for the half-reactions involved. The standard reduction potentials can be found in tables, such as the Standard Reduction Potentials table. In this case, we have Zn2+ + 2e- --> Zn, and Bi3+ + 3e- --> Bi. The standard reduction potentials for these reactions are -0.763 V and -0.507 V, respectively.
By using the Nernst equation, substituting the known values, and assuming standard conditions (T = 298 K, Q = 1), we can calculate the cell potential (Ecell):
Ecell = E°cell - (0.0257/n) * ln(Q)
b) 2H2 + O2 --> 2H2O (in acidic solution)
In an acidic solution, we need to balance the charges of the half-reactions by adding H+ and H2O as needed. The balanced half-reactions are:
O2 + 4H+ + 4e- --> 2H2O (reduction)
2H2O --> 4H+ + O2 + 4e- (oxidation)
You can look up the standard reduction potentials for these reactions and use the same process as in part (a) to calculate the cell potential and standard reaction Gibbs free energy.
c) 2H2 + O2 --> 2H2O (in basic solution)
In a basic solution, the main difference is the addition of OH- ions to the balanced half-reactions to neutralize the H+ ions. The balanced half-reactions become:
O2 + 4OH- + 4e- --> 2H2O + 4OH- (reduction)
2H2O + 2e- --> H2 + 2OH- (oxidation)
Again, you can determine the standard reduction potentials, plug them into the Nernst equation, and calculate the cell potential and standard reaction Gibbs free energy.
d) 3Au+ --> 2Au + Au3+
For this reaction, you can find the standard reduction potentials for the half-reactions involved (Au3+ + 3e- --> Au and Au+ + e- --> Au). Then you can use the same process as in the previous examples to calculate the cell potential and standard reaction Gibbs free energy.
Remember to make sure to use the correct values for the standard reduction potentials for your calculations. Additionally, be aware of any changes in stoichiometry or conditions (such as acidic or basic solutions) that may affect the reaction and the calculations.