Question

Give half cell equation of Daniel cell takes place at cathode.

Answer 1

In Daniel cell, reduction takes place at cathode:
$C{u}^{2+}\left(1.0M\right)+2e^{-}\to Cu\left(s\right)$
Oxidation occurs at anode.
$Zn\left(s\right)\to Z{n}^{2+}\left(1.0M\right)+2e^{-}$
The net cell reaction is
$Zn\left(s\right)+C{u}^{2+}\left(1.0M\right)\to Z{n}^{2+}\left(1.0M\right)+Cu\left(s\right)$