Question

The following cell has a potential of 0.040 V at 25°C. Zn(s)∣Zn2+(aq) || Cr3+(0.020 M) | Cr(s) What is the concentration of Zn2+? The standard reduction potentials are given below: Zn2+(aq) + 2e− → Zn(aq) Eo = − 0.760 Cr3+(aq) + 3e− → Cr(s) V Eo = − 0.740 V

Answer 1

from data table:

Eo(Zn2+/Zn(s)) = -0.760 V

Eo(Cr3+/Cr(s)) = -0.740 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Cr3+/Cr(s))

anode is (Zn2+/Zn(s))

The chemical reaction taking place is

2 Cr3+(aq) + 3 Zn(s) --> 2 Cr(s) + 3 Zn2+(aq)

Eocell = Eocathode - Eoanode

= (-0.740) - (-0.760)

= 0.020 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

we have below equation to be used:

E = Eo - (0.0592/n) log {[Zn2+]^3/[Cr3+]^2}

0.04 = 0.02 - (0.0592/6) log ([Zn2+]^3/0.02^2)

log ([Zn2+]^3/0.02^2) = -2.027

([Zn2+]^3/0.02^2) = 9.397*10^-3

[Zn2+]^3 = 3.759*10^-6

[Zn2+] = 1.56*10^-2 M

Answer: 1.56*10^-2 M