Question at position 31 Which of the following redox reactions will be spontaneous under standard conditions? I. Co2+(aq) + 2Fe2+(aq) Co(s) + 2Fe3+(aq)II. 3Co3+(aq) + Sn(s) 3Co2+(aq) + Sn2+(aq)III. I2(s) + 2 Br–(aq) 2 I–(aq) + Br2(l)IV. Mn(s) + Cl2(g) Mn2+(aq) + 2Cl–(aq) I II and IV II III II and III I and III IV单项选择题
A
I
B
II and IV
C
II
D
III
E
II and III
F
I and III
G
IV

Question

Question at position 31 Which of the following redox reactions will be spontaneous under standard conditions? I. Co2+(aq) + 2Fe2+(aq) Co(s) + 2Fe3+(aq)II. 3Co3+(aq) + Sn(s) 3Co2+(aq) + Sn2+(aq)III. I2(s) + 2 Br–(aq) 2 I–(aq) + Br2(l)IV. Mn(s) + Cl2(g) Mn2+(aq) + 2Cl–(aq) I II and IV II III II and III I and III IV单项选择题

A

I

B

II and IV

C

II

D

III

E

II and III

F

I and III

G

IV

BlackTom AI · Accepted Answer

Reduction potentials determine whether a redox reaction is spontaneous: E°cell = E°(reduction at cathode) − E°(reduction at anode). If E°cell > 0 the reaction is spontaneous under standard conditions. Use standard half‑cell potentials (V): Fe3+/Fe2+ = +0.77; Co2+/Co = −0.28; Co3+/Co2+ ≈ +1.82; Sn2+/Sn = −0.14; Br2/Br− ≈ +1.07; I2/I− = +0.54; Cl2/Cl− = +1.36; Mn2+/Mn = −1.18.

Step-by-step for each reaction:

I. Co2+ + 2Fe2+ → Co + 2Fe3+
- Reduction candidate: Co2+ + 2e− → Co, E°red = −0.28 V (cathode if reduced).
- Oxidation candidate: Fe2+ → Fe3+ + e− is reverse of Fe3+ + e− → Fe2+ (E°red = +0.77 V), so E°ox = −0.77 V.
- E°cell = E°cathode − E°anode = (−0.28) − (+0.77) = −1.05 V → negative, nonspontaneous.

II. 3Co3+ + Sn → 3Co2+ + Sn2+ (balanced relevant form is 2Co3+ + Sn → 2Co2+ + Sn2+)
- Reduction: Co3+ + e− → Co2+, E°red = +1.82 V.
- Oxidation: Sn → Sn2+ + 2e− is reverse of Sn2+ + 2e− → Sn, E°red = −0.14 V so E°ox = +0.14 V.
- E°cell = E°(Co3+/Co2+) − E°(Sn2+/Sn) = 1.82 − (−0.14) = 1.96 V (positive) → spontaneous. The stoichiometry in the problem (3 vs 2) does not change the sign of E°cell.

III. I2 + 2Br− → 2I− + Br2
- Reduction potentials: I2 + 2e− → 2I− E° = +0.54 V; Br2 + 2e− → 2Br− E° = +1.07 V.
- Bromine has the higher reduction potential, so Br2 is reduced and I− would be oxidized in a spontaneous cell. The written direction (I2 being reduced while Br− is oxidized) gives
- E°cell = E°(I2/I−) − E°(Br2/Br−) = 0.54 − 1.07 = −0.53 V → negative, nonspontaneous.

IV. Mn + Cl2 → Mn2+ + 2Cl−
- Reduction: Cl2 + 2e− → 2Cl−, E° = +1.36 V.
- Oxidation: Mn → Mn2+ + 2e−, reverse of Mn2+ + 2e− → Mn, E°red = −1.18 V.
- E°cell = 1.36 − (−1.18) = 2.54 V → strongly positive, spontaneous.

Why the correct choice is II and IV: both II and IV give positive E°cell values, so they proceed spontaneously under standard conditions. Reactions I and III give negative E°cell and are nonspontaneous as written. Therefore the correct answer is II and IV, which matches the provided answer.

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