There is considerable interest in producing ammonia by electrolyzing nitrogen/water mixtures (in the — Electrochemistry and Thermodynamics Chemistry Question
Problem Context
There is considerable interest in producing ammonia by electrolyzing nitrogen/water mixtures (in the presence of suitable catalysts) as shown below.
N2(g) + 3 H2O(l) → 2 NH3(g) + 1.5 O2(g)
Would ammonia be evolved at the cathode or the anode of this electrolytic cell?
Model Answer
Nitrogen is reduced to ammonia in this cell, so ammonia is produced at the cathode.
Calculate the standard cell potential Eº for this reaction at 298 K given the thermodynamic data shown.
Model Answer
∆Gº = 2(–16.4 kJ mol-1) – 3(–237.1 kJ mol-1) = 678.5 kJ mol-1
∆Gº = –nFEº
678500 J mol-1 = –(6)(96500 C mol-1)Eº
Eº = –1.17 V
Would Eº for this reaction be larger or smaller in magnitude at higher temperatures? Justify your answer.
Model Answer
In the reaction, 3.5 mol of gaseous products are produced from 2 mol gaseous reagents, so ∆Sº is positive (it turns out to be 100.2 J mol-1 K-1). Since ∆Gº = ∆Hº – T∆Sº, if ∆Sº is positive, ∆Gº will become algebraically smaller as temperature increases. Since ∆Gº is positive, this means that its magnitude will become smaller as T increases, and hence Eº will become smaller in magnitude (less negative) as T increases.
The standard reduction potential of O2 in acidic solution is 1.23 V.
O2(g) + 4 H+(aq) + 4e– → 2 H2O Eº = 1.23 V
Write the half-reaction corresponding to the reduction of O2(g) to water in basic solution and calculate the standard reduction potential.
Model Answer
In basic solution, O2(g) + 2 H2O(l) + 4e– → 4 OH–(aq)
To calculate Eº, one can combine the ∆Gº values for the reduction reaction in acid solution with the ∆Gº for the autoprotolysis of water:
O2(g) + 4 H+(aq) + 4e– → 2 H2O(l) ∆Gº = –nFEº = –475 kJ mol-1
4 H2O(l) → 4 H+(aq) + 4 OH–(aq) ∆Gº = –RTln(Keq) = –RTln([1 × 10-14]4) = 319 kJ mol-1
O2(g) + 2 H2O(l) + 4e– → 4 OH–(aq) ∆Gº = –156 kJ mol-1
Eº = ∆Gº/(–nF) = (–156000 J mol-1)/(–4F) = 0.40 V
Write the half-reaction corresponding to the reduction of N2(g) to NH3(g) in basic solution and calculate the standard reduction potential.
Model Answer
N2(g) + 6 H2O(l) + 6e– → 2 NH3(g) + 6 OH–(aq)
Since the difference in standard reduction potentials for the two half-reactions is equal to the overall Eº, then Eº – 0.40 V = –1.17 V, and Eº = –0.77 V.
A cell for forming NH3 by electrolysis of nitrogen as described above runs for 3000 s with a current of 1.20 A and produces 1.05 × 10-3 mol NH3. What is the faradaic yield of ammonia (that is, the yield of ammonia as a percentage of the maximum amount that could be formed by passage of this amount of electricity)?
Model Answer
Theoretical yield of NH3 = (3000 s)(1.2 A)/(3)(96500 C mol-1) = 0.0124 mol NH3. The faradaic yield is thus {(1.05 × 10-3 mol)/(0.0124 mol)}×100% = 8.4%.