Standard electrode potential of a galvanic half cell is zero at equilibrium at standard conditions? At standard conditions(at 1atm pressure and unit activity(1molal or 1molar concentration)of all dissolved compounds),the electrode potential is equal to the standard electrode potential by the Nernst equation. $$ \Delta E = \Delta E^0 $$
At the same time,while the electrochemical reaction is at equilibrium,the electrode potential of the cell, $$ \Delta E =0 $$ 

So this must mean,that when ANY reaction is at equilibrium under the standard conditions, $$ \Delta E = \Delta E^0 = 0 $$
Which is quite surprising.Is this correct?please explain.
There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential. 

$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what  concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$. 

Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.

For $\Delta E$ however, consider the Nernst equation, 
$$E = E^\circ - (RT/nF)\ln Q$$
Change in $E$
$$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$
therefore
$$\Delta E = -(RT/nF)\Delta\ln Q$$

At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero).