2 Al(s) +Cr2O72-(aq) + 14H+(aq) 2 Al3+(aq) +2 Cr3+(aq) + 7 H2O(l) I keep getting Q to equal zero after using theNernst equation? How to do this problem in general would begreat? Calculate the pH of the cathode compartment for the followingreaction given script ecell =3.01 V when [Cr3+] = 0.15 M, [Al3+]= 0.30 M, and [Cr2O72-] =0.55 M. 2 Al(s) +Cr2O72-(aq) + 14H+(aq) ? 2 Al3+(aq) +2 Cr3+(aq) + 7 H2O(l) I keep getting Q to equal zero after using theNernst equation? How to do this problem in general would begreat?
Explanation / Answer
We Know that : The given reaction is : 2 Al(s) +Cr2O72-(aq) + 14H+(aq) 2 Al3+(aq) +2 Cr3+(aq) + 7 H2O(l) Oxidation half cell: Al -------> Al+3 + 3e ] x 2 Reduction half cell : Cr2 O7-2 + 6e -------> 2 Cr+3 According to Nernest equation : Ecell = E0cell - 0.059 / 6 log [ Al+3 ]2 [Cr+3 ]2 / [ Cr2O7-2 ] 3.01 V = E0cell - 0.059 / 6 log [ 0.30M ]2 [0.15 M ]2 / [ 0.55 M ] E0cell = 3.01 + 0.059 / 6 x -2.433 = 2.98602 V The value of Q obtained is -2.433 = 2.98602 V The value of Q obtained is -2.433