For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and E = 0.78 V? Assume T is 298 K

The Fe ion concentration in the galvanic cell has been 0.0186 M.

The Nernst equation has been used for the determination of the voltage of the galvanic cell.

It can be given as:

where, n = number of electrons transfer = 2

= initial voltage = 0.77 V

= cell voltage = 0.78 V

Iron ion concentration = 0.040 M

Susbstituting the values:

0.78 = 0.77 -

0.78 - 0.77 = -0.0296

-0.03378 =

antilog (-0.03378) =

Iron ion concentration = 0.0186 M

Thus, the Fe ion concentration in the galvanic cell has been 0.0186 M.

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0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

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