1. Given the half reaction 1/8NO$_3^-$ + 5/4H$^+$ + e$^-$ $\leftrightarrow$ 1/8NH$_4^+$ + 3/8H$_2$O with log K = 14.90 and a water sample with a$_{NO_3^-}$ = 10$^{-5}$, a$_{NH_4^+}$ = 10$^{-3}$, pH=8, a = m, 25°C, 1bar. Based on the redox couple NO$_3^-$ - NH$_4^+$, calculate the Eh of the solution? Remember Eh = 0.0592(pe). Report your answer on Canvas with 3 digits after the decimal.