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