The emf(in V) of a Daniel cell containing $0.1MZnS{O}_4$ and  $0.01MCuS{O}_4$ solution at their respective electrodes is $\left(E_{C{u}^{2+}/Cu}^0=0.34v,E_{Z{n}^{2+}/Zn}^0=-0.76v\right)$ 

$E_{cell}=E_{Cell}^0-\frac{0.06}{n}\log \frac{\left[\text{products}\right]}{\left[\text{reactants}\right]}$

$E_{cell}^0=E_{Cathode}^0-E_{Anode}$

At Anode  $Zn\to Z{n}^{+2}+2\overline{c}$
At Cathode  $\underset{¯}{C{u}^{2+}+2\overline{c}\to Cu}$

              $\underset{¯}{C{u}^{2+}+zn\to Cu+Z{n}^{+2}}$

$E_{Cell}=0.34-\left(-0.76\right)-\frac{0.06}{2}\log \frac{{10}^{-1}}{{10}^{-2}}$
               $=1.07V$