Answer: A
Solution: A. According to the diagram, $\mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$ can oxidize $\mathrm { Mn } ^ { 2 + }$ to $\mathrm { MnO } _ { 4 } ^ { - } , \mathrm { MnO } _ { 4 } ^ { - }$ and $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$. INLINE_FORMULA_5]] can oxidize $\mathrm { Fe } ^ { 2 + }$, so $\mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - } , \mathrm { Fe } ^ { 2 + }$ can not coexist in large quantities in aqueous solution, so A is wrong. B. In reaction I, $\mathrm { Mn } ^ { 2 + }$ is converted into $\mathrm { MnO } _ { 4 } ^ { - }$, and the valence of manganese increases. INLINE_FORMULA_8]] is converted to $\mathrm { MnO } _ { 4 } ^ { - }$, the valence of manganese element rises and $\mathrm { Mn } ^ { 2 + }$ is a reducing agent in reaction I, so B is correct. C. When reaction I regenerates $\mathrm { Mn } ^ { 2 + }$ is oxidized to $\mathrm { MnO } _ { 4 } ^ { - } , \mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$, there are 2 -1 valences in $\mathrm { O } , \mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$, which is reduced to [[INLINE_FORMULA_12 13]] is reduced to $\mathrm { SO } _ { 4 } ^ { 2 - }$, and the -1-valent oxygen valence is reduced to $- 2,1 \mathrm {~mol} \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$ and 2 mol of electrons are transferred, according to the conservation of electrons, ${ } ^ { 2.5 \mathrm { molS } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - } }$ is consumed when regenerating $1 \mathrm { molMnO } _ { 4 } ^ { - }$, so C is correct. is correct;
D. The solution changes from colorless to purplish-red when reaction I occurs, and the purplish-red color of the solution fades when reaction II occurs, so D is correct; choice A. The solution changes from colorless to purplish-red when reaction I occurs.
$37 . \mathrm { C }$
[Knowledge Points]conceptual judgment of redox reactions, redox reaction definition, nature and characteristics, redox reaction related calculations, electron transfer calculations
[Analysis]In the reaction $\mathrm { NH } _ { 4 } \mathrm { NO } _ { 3 } + \mathrm { Zn } \xlongequal { \text { 温热 } } \mathrm { ZnO } + \mathrm { N } _ { 2 } \uparrow + 2 \mathrm { H } _ { 2 } \mathrm { O }$, $\mathrm { NH } _ { 4 } \mathrm { NO } _ { 3 }$, the N element is $- 3 , + 5$ valence, and after the reaction, they are all converted to 0-valence; Zn is formed into ZnO, and the 0-valence is raised to +2-valence, so answer accordingly.
A. The essence of redox reactions is the loss or gain of electrons or the shift of shared electron pairs, which can be judged by whether the elemental valence changes, by analyzing the reaction before and after the change in the elemental valence, so A does not meet the meaning of the question;
B. Nitrogen is generated by the change of the valence state of the N element in $\mathrm { NH } _ { 4 } \mathrm { NO } _ { 3 }$, and the valence state of the N element in $\mathrm { NH } _ { 4 } \mathrm { NO } _ { 3 }$ both increases and decreases, so nitrogen is both an oxidized product and a reduced product, so B does not meet the meaning of the question;
C. Under standard conditions, $11.2 \mathrm { LN } _ { 2 }$ is 0.5 mol, and the number of electrons outside the nucleus is equal to the atomic number, so the amount of electrons is $0.5 \mathrm {~mol} \times 7 \times 2 = 7 \mathrm {~mol}$, so C meets the question;
D. In the reaction, the number of electrons transferred is $5 \mathrm { e } ^ { - }$, so for every 1 mol of ZnO, a total of 5 mol of electrons are transferred, so D does not meet the meaning of the question;
Therefore, the answer is C.