Bloch electrons and their velocity
By Bloch's theorem, the single-electron energy eigenstates in a crystal (with periodic potential) are Bloch states of the form
$$\psi_{\mathbf k}(\mathbf r)=u(\mathbf r)e^{i\mathbf k\cdot\mathbf r}$$
where $u(\mathbf r)$ is a periodic function with the periodicity of the crystal and $\mathbf k$ is the wave vector. The Bloch states $\psi_{\mathbf k}(\mathbf r)$ are the single-particle states whose energy eigenvalues make up the nearly-continuous conduction band. If you have studied any quantum mechanics, ignoring the $u(\mathbf r)$ factor for a moment, you will recognize $\psi_{\mathbf k}(\mathbf r)$ as a momentum (and hence velocity) eigenfunction with momentum eigenvalue $\hbar \mathbf k$. However, Bloch states are not momentum (or velocity) eigenfunctions and $\hbar \mathbf k$ is not their momentum (this is called crystal momentum). Nonetheless, we can still compute the expected value of velocity in a Bloch state, and it turns out to be
$$\mathbf v(\mathbf k) =\frac{1}{\hbar}\nabla_\mathbf k E(\mathbf k)$$
where $E(\mathbf k)$ is the energy eigenvalue. You can verify that this also gives the correct velocity for a free particle. It might also look familiar from group velocity of waves, which is calculated from the dispersion relation as $\mathbf v(\mathbf k) =\nabla_\mathbf k \omega(\mathbf k)$.
To answer your question: since $E_C $ is the minimum of $E(\mathbf k)$ in the conduction band, and since we must have $\nabla_\mathbf k E(\mathbf k) = 0$ at the minimum, the Bloch electron with $E(\mathbf k)=E_C$ has zero mean velocity and thus cannot contribute to current. It is therefore reasonable to say that such an electron has zero kinetic energy (i.e. energy due to motion). However, it is potentially misleading to say that an electron with energy $E_C$ cannot contribute to current because such an electron can still accelerate under the influence of an external E-field, moving up in the conduction band and acquiring a non-zero velocity.
Parabolic bands
In many (most?) semiconductors, it turns out that the conduction band is parabolic. Mathematically, this means that for energies close to $E_C$,
$$E(\mathbf k) = E_C +\frac{\hbar^2}{2}(\mathbf k - \mathbf k_0)^T\mathbf M^{-1}(\mathbf k - \mathbf k_0).$$
Here, $\mathbf M^{-1}$ is called the inverse effective mass tensor (basically a symmetric $3\times 3$ matrix), and $\mathbf k$ and the constant $\mathbf k_0$ are now thought of as column vectors. Purely out of convenience I will assume $\mathbf k_0 = 0$, but you can just substitute $\hat{\mathbf{k}} = \mathbf k - \mathbf k_0$ for $\mathbf k$ in what follows to proceed without this assumption.
With an appropriate choice of coordinates, we can write the energy as
$$E(\mathbf k) = E_C +\frac{\hbar^2}{2}\left(\frac{k_x^2}{m^*_x}+\frac{k_y^2}{m^*_y}+\frac{k_z^2}{m^*_z}\right)$$
which explains the term "parabolic": the energy varies parabolically as you move in a straight line in $\mathbf k$-space. In an isotropic band where $m^*_x = m^*_y = m^*_z =m^*$, we have simply
$$E(\mathbf k) = E_C +\frac{\hbar^2k^2}{2m^*}$$
where the second term might be familiar as the (kinetic) energy eigenvalue of a free particle with mass $m^*$.
Velocity and kinetic energy in a parabolic band
The mean (group) velocity in a parabolic band is
$\mathbf v(\mathbf k) =\frac{1}{\hbar}\nabla_\mathbf k E(\mathbf k) = \hbar \mathbf M^{-1}_{ij}\mathbf k$, and the electron energy can be written as
$$E(\mathbf k) = E_C + \frac{1}{2}\mathbf v^T \mathbf M \mathbf v.$$
The second term is now easy to think of as kinetic energy (the energy due to motion). It is entirely analogous to the formula $\frac12mv^2$, and in fact reduces to it in an isotropic band.
