Question
for this equation
Fe(CN)6 4- + Fe 3+ --> Fe(CN)6 3- + Fe 2+
at 310 K and pH 2
using the equation
E = Eo + 2.303 RT/nF log K - 2.303 pH RTh/nF
what is h ?
Fe(CN)6 4- + Fe 3+ --> Fe(CN)6 3- + Fe 2+
at 310 K and pH 2
using the equation
E = Eo + 2.303 RT/nF log K - 2.303 pH RTh/nF
what is h ?
Answers
bobpursley
Isn't h the natural log of the reaction quotient?
lucy
what so for the last part of the equation -2.303 RT/nF ln pH?