Question

The voltage of the cell below is 0.050 V at 25˚C. Determine the concentration of Zn2+ at the
cathode.
??(?)| ??
2+(??. 0.10?)|| ??2+ (??. ? ?)|??(?)

Answer 1

Given that:

E_cell = 0.050 V

[Zn²⁺]_anode = 0.10M

[Zn²⁺]_cathode = ?

At anode, oxidation takes place and at cathode, reduction takes place

In this reaction, number of electrons involved(n) are 2

NOw, from nernest equation

Ecell = (-0.0592/n) log[Zn²⁺]_oxidation / [Zn²⁺]_reduction

0.050 V = (-0.0592/2) log( 0.10/ [Zn²⁺]_reduction)

0.050 V*2/(-0.0592) =  log( 0.10/ [Zn²⁺]_reduction)

log( 0.10/ [Zn²⁺]_reduction) = -1.689

0.10/ [Zn²⁺]_reduction = antilog(-1.689) =  0.0205

[Zn²⁺]_reduction = 0.10/0.0205 = 4.878M

THus, concentration at cathode is 4.878 M