For the reaction: HCI(aQ) + Mn_3N_2(l) + H_2SO_4(aq) rightarrow S + MnCl_2(l) +
ID: 523829 • Letter: F
Question
For the reaction: HCI(aQ) + Mn_3N_2(l) + H_2SO_4(aq) rightarrow S + MnCl_2(l) + H_2O(l) + H_3N(aq) Write and label the oxidation half reaction. Write and label the reduction half reaction. Determine the oxidizing agent in this reaction. What are the anode and cathode of this reaction? List the formulas of the spectator ions. Balance the equation. If this reaction has a standard voltage of (E degree) of +5.6 mV, what is the standard Gibb's free energy of the reaction? Write the equilibrium expression for this reaction. From the value determined in (6), find the value of the equilibrium expression at 25 degree C.Explanation / Answer
1) The oxidation half is the oxidation of Mn3N2 to MnCl5.Both the compounds are electrically neutral and we need to determine the oxidation number of Mn in both the compounds.
The oxidation number of N is Mn3N2 is -3 and the oxidation number of Cl in MnCl5 is -1. Let x1 and x2 be the oxidation numbers of Mn in the two compounds.
Therefore,
3*(x1) + 2*(-3) = 0
===> 3x1 – 6 = 0
===> 3x1 = +6
===> x1 = +6/3 = +2
1*(x2) + 5*(-1) = 0
===> x2 – 5 = 0
===> x2 = +5
The oxidation numbers of Mn in Mn3N2 and MnCl5 are +2 and +5 respectively and the change in oxidation number is (+5) – (+2) = +3. The oxidation half reaction is:
Oxidation: Mn+2 -----> Mn+5 + 3 e-
2) The reduction half is the conversion of H2SO4 to S8. Both the compounds are electrically neutral and we have to ascertain the oxidation number of S in the compounds.
The oxidation numbers of H and O are +1 and -2. Let x1 and x2 be the oxidation numbers of S in the two compounds.
Therefore,
2*(+1) + 1*(x1) + 4*(-2) = 0
===> 2 + x1 – 8 = 0
===> x1 – 6 = 0
===> x1 = +6
8*(x2) = 0
===> x2 = 0
The oxidation numbers of S in the two compounds are +6 and 0. The reduction half reaction is
Reduction: 8 H2SO4 + 48 H+ + 48 e- -------> S8 + 32 H2O
3) The oxidizing agent is the one which itself undergoes a reduction. Since H2SO4 is reduced, H2SO4 is the oxidizing agent.
4) The cathode is where reduction takes place. Since H2SO4 is reduced to S8, the cathode is composed of H2SO4/S8. Offcourse, we shall need a Pt support to support the reducing agent and we can represent the cathode as
Cathode: PtH2SO4 (aq), S8 (g)
The anode is where the oxidation occurs. Since Mn3N2 is oxidized to MnCl5, the anode is Mn3N2/MnCl5. Again, we shall employ Pt as a solid support. The anode is
Anode: PtMn3N2 (aq), MnCl5 (aq)
5) The spectator ions are those which do not have their oxidation numbers changed in the reaction. Starting from the left, the spectator ions are H+, Cl-, N3-, O2-.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.