Question

If the concentration of $Z{n}^{2+}$ ions are 0.2 M and concentration of $C{u}^{+2}$ ions are 0.1 M respectively around their electrodes, the emf of the cell using $Cu$ and $Zn$ electrodes is:

• A. 0.90V
• B. 1.0771V
• C. 1.117V
• D. 0.076V

Answer 1

The correct option is B 1.0771V

Now, first solving for $E_{C{u}^{2+}|Cu}^0=?$

$-nF{E}_{C{u}^{2+}|Cu}^0=-nF{E}_{C{u}^{2+}|C{u}^{+}}^0-nF{E}_{C{u}^{+}|Cu}^0$

$2\times E_{C{u}^{2+}|Cu}^0=E_{C{u}^{2+}|C{u}^{+}}^0+E_{C{u}^{+}|Cu}^0$

$E_{C{u}^{2+}|Cu}^0=\frac{E_{C{u}^{2+}|C{u}^{+}}^0+E_{C{u}^{+}|Cu}^0}{2}=\frac{0.15+0.5}{2}=0.325V$

The standard emf of cell is

$E_{cell}^0=E_{C{u}^{2+}|Cu}^0-E_{Z{n}^{2+}|Zn+}^0=0.325V-(-0.76V)=1.085V$

The emf of cell is

$E_{cell}=E_{cell}^0-\frac{0.0592}{2}log\frac{\left[{Zn}^{2+}\right]}{\left[{Cu}^{2+}\right]}=1.085-\frac{0.0592}{2}log\frac{\left[0.2\right]}{\left[0.1\right]}=1.0771V.$