Question

# Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2...

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

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)red = -0.45

Zn(s) -------> Zn2+(aq) + 2e-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

Ecell= E°cell - (RT/nF) ln Q

Ecell = 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

Ecell = 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

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