Question

which statement among a - c is false regarding the redox reactions shown? electron acceptor electron donor for no2 - + 8h+ + 6e - → nh4+ + 2h2o so42 - + 10h+ + 8e - → h2s + 4h2o (mv)a - 440 - 220 δg° (kj) - 255 + 170 it requires energy to reduce sulfate (so42 - ). hydrogen sulfide (h2s) would be considered a strong electron donor. nitrite (no2 - ) is a stronger terminal electron acceptor than sulfate (so42 - ). none are false; all are true statements.

Explanation:

Step1: Analyze electron - donor strength

Hydrogen sulfide ($H_2S$) has a negative standard reduction potential ($E^0$) for its oxidation reaction ($H_2S + 4H_2O
ightarrow SO_4^{2 -}+10H^++8e^-$ with $E^0=- 0.220$ V), which indicates it can act as an electron - donor. A more negative $E^0$ value for an oxidation reaction implies a stronger electron - donating ability.

Step2: Analyze electron - acceptor strength

Nitrite ($NO_2^-$) has a positive standard reduction potential ($E^0 = + 0.440$ V) for its reduction reaction ($NO_2^-+8H^++6e^-
ightarrow NH_4^++2H_2O$), and sulfate ($SO_4^{2 -}$) has a less positive or more negative $E^0$ value compared to nitrite for its reduction process. A more positive $E^0$ for a reduction reaction implies a stronger electron - accepting ability. So nitrite is a stronger terminal electron acceptor than sulfate.

Step3: Analyze energy requirement for sulfate reduction

The reduction of sulfate ($SO_4^{2 -}$) is an endergonic process as indicated by the positive $\Delta G^0$ value ($\Delta G^0= + 170$ kJ), which means it requires energy to reduce sulfate.

Answer:

None are false; all are true statements.