1 mol/l 1 mol/l 1 mol/l

Its essence is still NO3-+3h+2e-= = HNO 2+H2O.

10-14 mol/l

Therefore E0(NO3-/NO2-)= E(NO3-/HNO2).

= E0(NO3-/HNO 2)+0.059 1v/2 * LG {([no3-]/c0)*([h+]/c0)^3/[HNO 2]/c0 }

= 0.94v+0.059 1/2v * LG( 10^- 14)^3

=-0.30 volts