The following electrodes and solutions are available for an experiment at 25 deg

Question

The following electrodes and solutions are available for an experiment at 25 degree C. Al, Zn, KNO_3 (0.1 M), Al(NO_3)_3 (1.0 M), and Mn(NO_3)_2 (1.0 M) Complete the following for a ELECTROLYTIC cell: This reaction is not spontaneous because E degree is Delta G is and K is The equilibrium lies with the Calculate the new E_cell if 0.1 M Al(NO_3)_3 was used (all other parameters remain the same). Calculate the new E_cell if 2.0 M Al(NO_3)_3 was used (all other parameters remain the same). Generalizations for the Nernst Equation:If the concentration of the products decrease in a electrolytic cell, E If the concentration of the products increase in a electrolytic cell, E

Explanation / Answer

2.

Anode = Al3+/Al / Half-cell reaction : Al ---> Al3+ + 3e-

Cathode = Zn2+/Zn / Half-cell reaction : Zn2+ + 2e- ---> Zn

Balanced cell equation : 2Al + 3Zn2+ ---> 2Al3+ + 3Zn

Cell diagram : Al(s) | Al(NO3)3 (1.0 M) || Zn(NO3)2 (1.0 M) | Zn

Eo = -0.763 - (-1.66) = 0.897 V

dGo = -nFEo = -6 x 96485 x 0.897 = -519.28 kJ

K = inv.ln(519282.3/8.314 x 298) = 1.06 x 10^91

This reaction is spontaneous because, Eo is +ve, dGo is -ve, K is large. The equilibrium lies with the products.

When, Al(NO3)3 = 0.1 M

Ecell = 0.897 - 0.0592/6 log(0.1^2/1^3) = 0.917 V

If Al(NO3)3 = 2.0 M

Ecell = 0.897 - 0.0592/6 log(2^2/1^3) = 0.891 V

Generalisation of Nernst equation,

If concentration of reactants decreases in the cell, E = low

If concentration of reactant increase in a cell, E = large

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.