Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.837 M and [Fe2 ] = 0.0100 M. Standard reduction potentials can be found here.

Zn(s)+Fe^2+(aq) <--->Zn^2+(aq)+Fe(s)

E= _______ V

Answer 1

From the reaction given it is clear that zinc is oxidized i.e. electron are losed by zinc . Iron is reduced i.e. it gains electrons

From the reaction above lets write half reactions

Fe⁺²_(aq) + 2e^- -------> Fe _(s) E°_red = -0.45

Zn_(s) -------> Zn²⁺_(aq) + 2e^- E°_oxd = +0.76

Net reaction - Zn_(s) + Fe⁺²_(aq) -----> Zn⁺²_(aq) + Fe_(s) E°​_cell = 0.31

Now will we use Nernst equation

E_cell= E^°_cell - (RT/nF) ln Q

E_cell = E^°_cell - (RT/nF) ln ([Zn²⁺] / [Fe²⁺])

Here

E^°_cell = 0.31

R = 8.314 ( gas constant )

n = no. Of electrons transferred = 2

F = 96500 columbs

[Zn²⁺] = 0.837 M

[Fe²⁺] = 0.0100 M

T = 25^° c = 273 +25 = 298 k

Substitute

E_cell = 0.31 - ((8.314*298)/(2*96500)) ln (0.837/0.0100)

= 0.31 - (2477.572/193000) ln (87)

= 0.31 - 2.303 * 0.0128 * log (87) (2.303 is multiplied to remove the natural log ln )

= 0.31 - 2.303 * 0.0128 * 1.9395

= 0.31 - 0.0571

= 0.2529 V

Hence E = 0.2529 V