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.

For more information about the galvanic cell, refer to the link;

link

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

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.