2 Al( s ) +Cr 2 O 7 2- ( aq ) + 14H + ( aq ) 2 Al 3+ ( aq ) +2 Cr 3+ ( aq ) + 7

Question

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

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.