Is the spin 1/2 rotation matrix taken to be counterclockwise? The spin 1/2 rotation matrix around the $z$-axis I worked out to be
$$
e^{i\theta S_z}=\begin{pmatrix}
\exp\frac{i\theta}{2}&0\\
0&\exp\frac{-i\theta}{2}\\
\end{pmatrix}
$$
Is this taken to be anti-clockwise around the $z$-axis?
 A: For your example, we have $e^{i\theta S_z}\mathbf{S}e^{-i\theta S_z}=\begin{pmatrix}\cos\theta & -\sin\theta&0\\\sin\theta & \cos\theta&0\\0&0&1\end{pmatrix}\mathbf{S}$, with $e^{i\theta  S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}$ and $\mathbf{S}=\begin{pmatrix}S_x\\ S_y\\ S_z\end{pmatrix}$ representing the spin-1/2 operators. 
Comments: 
In fact, for the most general spin rotation, we have $$U\mathbf{S}U^\dagger=A\mathbf{S}\rightarrow (1)$$, where $U$ represents the general spin rotation operator
$U=e^{i\alpha  S_z}e^{i\beta  S_y}e^{i\gamma S_z}=\begin{pmatrix}\cos{\frac{\beta }{2}}e^{i\frac{\alpha + \gamma}{2}} & \sin{\frac{\beta }{2}}e^{i\frac{\alpha - \gamma}{2}}\\ -\sin{\frac{\beta }{2}}e^{i\frac{\gamma-\alpha}{2}} & \cos{\frac{\beta }{2}}e^{-i\frac{\alpha + \gamma}{2}}\end{pmatrix}\in SU(2)$, and 
$A=\begin{pmatrix}\cos\alpha \cos\beta\cos\gamma-\sin\alpha\sin\gamma& -\sin\alpha \cos\beta\cos\gamma-\cos\alpha\sin\gamma &\sin\beta\cos\gamma\\ \cos\alpha \cos\beta\sin\gamma+\sin\alpha\cos\gamma &  -\sin\alpha \cos\beta\sin\gamma+\cos\alpha\cos\gamma&\sin\beta\sin\gamma\\-\cos\alpha\sin\beta&\sin\alpha\sin\beta&\cos\beta\end{pmatrix}$ $\in SO(3)$ with the three Euler angles $\alpha,\beta,\gamma$.
Eq.(1) gives the map from $SU(2)$ to $SO(3)$ and the relation $SO(3)\cong SU(2)/Z_2.$
Remarks:
$e^{i\theta  S_x}=\begin{pmatrix}\cos{\frac{\theta }{2}} & i\sin{\frac{\theta }{2}}\\ i\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}}
\end{pmatrix},e^{i\theta  S_y}=\begin{pmatrix}\cos{\frac{\theta }{2}} & \sin{\frac{\theta }{2}}\\ -\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}}\end{pmatrix},e^{i\theta  S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}.$
A: The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by:
$R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right)
  ~=~ I\cos\frac{\phi}{2}+\left(-i\,\hat{s}\cdot\vec{\sigma}\right)\sin\frac{\phi}{2}$
Where $\vec{\sigma}=\{\sigma_x,\sigma_y,\sigma_z\}$ and $I$ is the unit matrix which is the same as $\sigma_o$. We can explicitly write the generators of (right-handed) rotation as follows starting from the definition of the Pauli matrices.
:\begin{align}
  \sigma_x =
    \begin{pmatrix}
      ~~0&~~1\\
      ~~1&~~0~~
    \end{pmatrix} &&
  \sigma_y =
    \begin{pmatrix}
      ~~0&-i\\
      ~~i&~~0~~
    \end{pmatrix} &&
  \sigma_z =
    \begin{pmatrix}
      ~~1&~~0\\
      ~~0&-1~~
    \end{pmatrix} \,
\end{align}
:\begin{align}
  j_x = 
    \begin{pmatrix}
      ~~0&-i\\
      -i&~~0~~
    \end{pmatrix} &&
  j_y = 
    \begin{pmatrix}
      ~~0&-1\\
      ~~1&~~0~~
    \end{pmatrix} &&
  j_z = 
    \begin{pmatrix}
      -i&~~0\\
      ~~0&~~i~~
    \end{pmatrix} \,
\end{align}
The specific rotation matrix as given in the question above is a left-handed rotation since the right-handed rotation matrix is defined by:
$R~=~ \exp\left(\frac{\phi}{2} j_z\right)
  ~=~ I\cos\frac{\phi}{2}\phi~+~j_z\sin\frac{\phi}{2}
  ~=~ \begin{pmatrix}
      \exp-i\frac{\phi}{2}&0\\
      ~~0&\exp i\frac{\phi}{2}~~
    \end{pmatrix}$
Counter clockwise is right-handed if the rotation axis points towards you, but it is left-handed if the rotation-axis points away from you. It's up to your choice...
