Tutorial 8: Sulfur Oxidation States and Copper Redox Reactions, Study notes of Chemical Principles

Download Tutorial 8: Sulfur Oxidation States and Copper Redox Reactions and more Study notes Chemical Principles in PDF only on Docsity! CM 1111 Tutorial 8 Question 1 Shown below is a Latimer diagram for sulfur oxoanions determined at pH = 0. SO4 2- S2O6 2- SO2 Eo = X Eo = +0.40 VEo = +0.57 VEo = –0.25 V S2O3 2- S8 Eo = + 0.5 V Eo = Y a) Find the standard reduction potentials X and Y. First, find the oxidation states of the sulfur atoms in the molecules, because it will tell you the number of electrons involved in each reduction step: SO4 2- S2O6 2- SO2 Eo = X Eo = +0.40 VEo = + 0.57 VEo = -0.25 V S2O3 2- S8 Eo = + 0.5 V Eo = Y +1e - 0 +1e - +2e - +2e - +VI +V +IV +II X = [1 x (-0.25) + 1 x (+0.57)] / 2 = +0.16 V 0.5 = [2 x (0.4) + 2(Y)] / 4; Y = 0.6 V b) Is the disproportionation of S2O6 2– (reaction 1) thermodynamically favorable? Briefly explain your answer (hint: protons are needed to balance the half-reactions involved in the overall disproportionation). S2O6 2– SO4 2– + SO2 (reaction 1) To evaluate if the disproportionation is favorable or not, one needs to find the ∆E of the following half reactions: S2O6 2– + 4 H+ + 2e– 2 SO2 + 2 H2O E1 = 0.57 V 2 SO4 2– + 4 H+ + 2e– S2O6 2– + 2 H2O E2 = -0.25 V 2 S2O6 2– 2 SO4 2– + 2 SO2 ∆E = +0.82 V ∆E for the disproportionation = E1 – E2 = +0.82 V Since ∆G = -nF∆E, the ∆G is negative, hence the disproportionation is thermodynamically favorable. (c) Combustion of sulfur-containing coal and petroleum produces sulfur dioxide SO2. It is believed that SO2 is responsible for ‘acid rain” which contains H2SO4. Explain why the formation of H2SO4 form SO2 is thermodynamically favorable. (Hint: look up the reduction potential of oxygen). The formation of H2SO4 from SO2 involves oxidation of sulfur from oxidation state +IV in SO2 to oxidation state +VI in H2SO4. Oxygen is the only oxidizing agent in air. To prove that oxidation SO2 by O2 is thermodynamic favorable, one needs to calculate the ∆E of the following reaction: O2 + 4 H+ + 4e– 2 H2O E1 2 H2SO4 + 4 H+ + 4e– 2 SO2 + 4 H2O E2 Overall reaction: O2 + 2 SO2 + 2 H2O 2 H2SO4 ∆E = E1 – E2 From part (a) of the question, you know E2 = 0.16 V From lecture notes or other sources, you know E1 = 1.23 V So ∆E = 1.23 - 0.16 = 1.07 V Since ∆G = -nF∆E, the ∆G is negative and hence the oxidation is thermodynamically favorable.