Oxidation-reduction equations disproportionation reactions

Unfortunately, most of the M-HNO complexes are insoluble in water, in contrast with MbnHNO, which shows to be remarkably inert toward the release of UNO (hour timescale). The latter fact supports the proposal of UNO being a long-lived intermediate in the disproportionation reaction of NH2OH catalyzed by pentacyanofer-rates, leading to bound NO+ as the final oxidized product,67 or in the six-electron reduction of SNP with 1,4-dimethylhydrazine (Equation 7.15).42 On the basis of crystallographic observation of some intermediates and DFT calculations, UNO has been proposed as a necessary intermediate in the six-electron reduction of N02 to NH3 catalyzed by cytochrome c nitrite reductase.68... 

Equation) analysis, an intrinsic barrier of 3kcalmoH was obtained for the 17-electron rhenium complexes.The facile oxidation and reduction of 17-electron complexes makes them quite susceptible to disproportionation reactions ... 

Redox reactions Oxidation states Reduction potentials Nernst equation Disproportionation Potential diagrams Frost-Ebsworth diagrams... 

The decomposition of hydrogen peroxide, shown in the preceding equation, is an example of a disproportionation reaction, in which one element undergoes both oxidation and reduction. In the ca.se of H2O2, the oxidation number of O is initially — 1. In the products of the decomposition, O has an oxidation number of —2 in H2O and of 0 in O2. 

Strategy First (1) set up the Nemst equation for the reduction half-reaction and calculate red. Then (2) repeat the calculation for the oxidation half-reaction, finding Eox. Finally (3), add rcd + ra if the sum is positive, disproportionation should occur. 

Studies of oxygen isotope fractionation were undertaken to address the mechanisms of the oxidative and reductive phases of SOD catalysis.44 Experiments were conducted at pH 10 in either borate or carbonate buffer where the rate is only slightly diminished from that at physiological pH. Control experiments demonstrated that all of the 02 produced came from the enzymatic reaction rather than the spontaneous disproportionation that occurs in the presence of trace metal ions slowly as the pH is increased. The reaction proceeds rapidly to 100% completion giving equal concentrations of 02 and H202 therefore, a ratio of fractionation factors was determined from the distribution of the lsO isotope in the two products. The ratio of fractionation factors reflects the ratio of KIEs on the oxidative and reductive reactions, and is designated as (3 (Equation 9.9). The (3 was determined to be 1.0104 0.0012 under the conditions described above. 

If further evidence of the difficulty of reducing the P=0 bond were needed, the recent findings of Yamashita et al ° provide interesting confirmation. Many a-hydroxyalkyldiphenylphosphine oxides (13) were converted into the corresponding alkyldiphenylphosphine oxides (14) in 33-73% yield, with PCb-KI (equation 21). No competing reduction of the P==0 bond was apparently observed. Majew-ski has reported the disproportionation of dialkylphosphine oxides into dialkyIphosphines and phos-phinic acids (equation 22). The reaction proceeds in yields of over 90% and the phosphines are easily isolated by distillation. 

The individual steps of this overall reduction reaction produce HCIO2, HOCl, and CI2, which all behave as oxidizing agents. An acidic medium is required, as CIO2 disproportionates in alkaline solution, as shown in equation 6. 

In certain cases, a given species acts simultaneously as both an oxidant in one of the couples and a reductant in the other couple. This reaction is called dismutation (or disproportionation). To balance such a reaction, one must first spot the fact that it is a case of dismutation, i.e., a particular chemical reaction involving two different redox couples. For example, the following equation may indeed be formally balanced, yet it does not correctly represent the chemical dismutation of diiodine because electrons appear on the right side, as in a redox half-reaction ... 

All the acceptors which are known to work in this scheme undergo irreversible side reactions following reduction e.q. disproportionation or decomposition. Thus, photolysis of Ru(bpy) with any of the acceptors listed in Table V in slightly acidic conditions leads to irreversible formation of RuCbpy) ". In the presence of suitable redox catalysts, RuCtpy) " efficiently oxidizes H O to 0, according to equation (52)... 

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