The following redox reaction is conducted with [Al3+] = 0.12 M and [Mn2+] = 1.5 M.

2 Al (s) + 3 Mn2 + (aq) → 2 Al3 + (aq) + 3 Mn (s) Eocell = 0.48 V Determine the moles of electrons transferred for the reaction as written (n), Q, and the cell potential (Ecell) at 298 K.

0.47V

Explanation:

2 Al (s) + 3 Mn2 + (aq) → 2 Al3 + (aq) + 3 Mn (s)

n = 6 (six moles of electrons were transferred)

Q = [Red]/[Ox] but [Red] = 1.5M, [Ox] = 0.12 M

Q = 1.5/0.12 = 12.5

From Nernst equation:

E = E°cell - 0.0592/n log Q

E°cell = 0.48 V

E = 0.48 - 0.0592/6 log (12.5)

E = 0.47V