Answer:
The equations are:
3Mn + 2NO3- + 8H+ -----------> 3Mn2+ + 2NO + 2H2O
2Mn2+ + 5IO4- + 3H2O ---------> 2MnO4- + 5IO3-+ 6H
Ecell = 1.6 V - 1.51 V = 0.09 V
Using the equation of Nernst ,
ΔG = -nFEcell
where n= number of electrons transferred = 10 mol
F = Farraday's constant = 96485 C/mol
Ecell = 0.09V
Now, ΔG = -10 * 96485 C/mol * 0.09 V
ΔG = -86836.5 J/mol (\because C*V =J)
ΔG = -86.84 kJ/mol
Now, Ecell = (0.0591/n)*log K
0.09 = 0.0591/10 *logK
0.09 = 0.00591 * log K
log K = 0.09 / 0.00591
log K = 15.23
K = 10-15.23
K = 5.89*10-16
The combined cell potential, gibbs free energy and value of K are; E°cell = 0.09 V; ΔG = -86.84 kJ/mol; K = 1.698 × 10¹⁴
A) The balanced equation describing each of the given reactions are;
3Mn + 2NO₃⁻ + 8H⁺ -----------> 3Mn²⁺ + 2NO + 2H₂O
2Mn²⁺ + 5IO⁴⁻ + 3H₂O ---------> 2MnO₄⁻ + 5IO₃⁻ + 6H
B) The combined cell potential is; E°cell = 1.6 V - 1.51 V
E°cell = 0.09 V
With the aid of the Nernst equation, we can get the gibbs free energy as;
ΔG = -nFE°cell
where;
n is number of electrons transferred = 10 mol
F is Faraday's constant = 96485 C/mol
E°cell = 0.09V
Thus,
ΔG = -10 * 96485 C/mol * 0.09 V
ΔG = -86836.5 J/mol
ΔG = -86.84 kJ/mol
Now, E°cell = (0.0591/n)*log K
0.09 = (0.0591/10) × log K
0.09 = 0.00591 × log K
log K = 0.09/0.00591
log K = 15.23
K = 1.698 × 10¹⁴
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