Question

Calculate the emf of the cell. $Zn|Z{n}^{2+}\left(0.001M\right)||C{u}^{2+}\left(0.1M\right)|Cu$
The standard potential of $Cu/C{u}^{2+}$ half-cell is +0.34 and $Zn/Z{n}^{2+}$ is -0.76 V.

Answer 1

Given $Zn|Z{n}^{2+}\left(0.001M\right)||C{u}^{2+}\left(0.1M\right)|Cu$

Overall cell reaction:

$Zn⟶{Zn}^{2+}+2e^{-}\underset{–}{{Cu}^{2+}+2e^{-}⟶Cu}\underset{–}{Zn+{Cu}^{2+}⟶{Zn}^{2+}+Cu}$

$E_{cell}^o=$standard reduction potential of cathode + standard oxidation potential of anode

$E_{cell}^o=$0.34 to 0.76 V

$E_{cell}^o=$1.1 V

$K_C=\frac{\left[{Zn}^{2+}\right]}{\left[{Cu}^{2+}\right]}=\frac{{10}^{-3}}{{10}^{-1}}={10}^{-2}$

EMF of the cell at any electrode concentration is:

$E=E^o-\frac{0.059}{n}\log \left(K_C\right)=1.1-\frac{0.059}{2}\log \left({10}^{-2}\right)=1.1-\frac{0.059}{2}\times \left(2\right)=1.1-0.059=1.041V$