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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.

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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

adioabiola

The no. of moles of electrons transferred is; n = 6.

The value of Q is; 12.5.

The cell potential (Ecell) at 298 K is; 0.471 V

The Redox reaction given is;

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

Evidently, n= 6 ( since, six moles of electrons were transferred from Mn²+ to Al)

  • Q= [Red]/[Ox]

pppwhere; [Red]= 1.5M, [Ox] = 0.12 M

  • Q= 1.5/0.12= 12.5

Ultimately, 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= 0.471 V

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