Some inorganic compounds exhibit a variety of oxidation states, for example, many Mn compounds are k — Organic Chemistry Chemistry Question
Electrochemistry of Inorganic Compounds
Some inorganic compounds exhibit a variety of oxidation states, for example, many Mn compounds are known with oxidation states ranging from 0 to VII. The standard reduction potential of a half reaction is measured against the hydrogen electrode. In this problem, the reduction Mn 2+ + 2 e – → Mn, E° = –1.18 V is expressed as Mn 2+ (–1.18) Mn.
For Mn in acidic solution the reduction series: Mn 3+ → Mn 2+ → Mn can be represented as follows: Mn 3+ (1.5) Mn 2+ (–1.18) Mn
A redox reaction takes place spontaneously if the redox potential is positive. A Frost diagram, a plot of nE° (n is the number of electron transferred in the half reaction) of the reduction couple X(N) / X(0) against the oxidation number, N, of the element, is used to indicate the most stable species of the compounds in different oxidation states. The Frost diagram of Mn 3+ / Mn 2+ / Mn is shown below.
The reduction potential depends on the concentration of the species in solution. The Ksp of MnCO3 is 1.8×10–11 . Use the Nernst equation to determine the potential at 25°C for the voltaic cell consisting of Mn(s) | Mn 2+ (aq) (1 M) || Mn 2+ (aq) / MnCO3 | Mn(s), if the concentration of Mn 2+ in the right hand side of the cell is 1.0×10–8 mol dm –3 .
Model Answer
For the concentration cell: Mn(s) | Mn 2+ (aq) (1M) || Mn 2+ (aq) / MnCO3 | Mn(s), (M = mol dm –3 )
Ecell = E o – (0.0592 / 2) log ([Mn 2+ ]right / [Mn 2+ ]left)
Ksp = 1.8×10 –11 = [Mn 2+ ][CO3 2– ]
[Mn 2+ ]right = 1.0×10 –8 and [Mn 2+ ]left = 1.0 with E o = 0.0 V (both are Mn)
Ecell = 0.0 – (0.0592 / 2) log (1.0×10 –8 / 1.0 ) = 0.237 V
For oxygen, the standard reduction potential in acidic solution can be represented as: O2 (0.70) H2O2 (1.76) H2O. Construct the Frost diagram for oxygen, and use the information contained in the diagram to calculate the reduction potential of the half reaction for reduction of O2 to H2O. Could H2O2 undergo disproportionation spontaneously?
Model Answer
Reduction of O2 to H2O is obtained as (0.70 V + 1.76 V) / 2 = 1.23 V, for O2 + 4 H + + 4 e– → 2 H2O E o = 1.23 V
The E o value could be obtained directly from the diagram by dividing the differences (2.46) of O2 and H2O by the differences of the oxidation number (2).
For H2O2 → O2 + H2O E o = 1.06 > 0.0
The disproportionation reaction is spontaneous.
Xenon difluoride can be made by placing a vigorously dried flask containing xenon gas and fluorine gas in the sunlight. The half–reaction for the reduction of XeF2 is shown below:
XeF2(aq) + 2 H + (aq) + 2e– → Xe(g) + 2 HF(aq) E° = 2.32 V
Use the VSEPR model to predict the number of electron–pairs and molecular geometry of XeF2. Show that XeF2 decomposes in aqueous solution, producing O2, and calculate the E o for the reaction. Would you expect this decomposition to be favoured in an acidic or a basic solution? Explain.
2 H2O → O2 + 4 H + + 4 e– E o = –1.23 V
Model Answer
The number of electron pair should be 5 (trigonal bipyramidal) with three electron pairs in the equatorial plane, thus the molecular geometry of XeF2 is linear.
2 H2O O2 + 4 H + + 4 e– E o = –1.23 V
XeF2(aq) + 2 H + (aq) + 2 e– Xe(g) + 2 HF(aq) E° = 2.32 V
2 XeF2(aq) + 2 H + (aq) + 2 H2O 2 Xe(g) + O2 + 4 HF(aq) E° = 1.09 V
The decomposition of XeF2 in aqueous solution is favoured in acidic solution.