The main results of the paper involve vanishing theorems for the Leibniz cohomology of semi-simple left Leibniz algebras over a field of characteristic zero. Recall that a left Leibniz algebra \(\mathfrak{L}\) (over a field \(k\)) is a \(k\)-vector space equipped with a bilinear map \(\circ :\mathfrak{L}\times\mathfrak{L}\to\mathfrak{L}\) that is a derivation from the left, namely \[x\circ(y\circ z)=(x\circ y)\circ z+y\circ (x\circ z), \] which is the mirror image of a right derivation for right Leibniz algebras introduced and studied in [J.-L. Loday and T. Pirashvili, Math. Ann. 296, No. 1, 139–158 (1993; Zbl 0821.17022)]. Let Leib\((\mathfrak{L})\) denote the vector subspace of \(\mathfrak{L}\) consisting of all linear combinations of squares \(x^2 :=x\circ x\), \(x\in\mathfrak{L}\). Then Leib\((\mathfrak{L})\) is an ideal of \(\mathfrak{L}\) and \(\mathfrak{L}/\mathrm{Leib}(\mathfrak{L}):=\mathfrak{L}_{\mathrm{Lie}}\) is a Lie algebra with bracket induced by the derivation operation \(\circ\). The Leibniz algebra \(\mathfrak{L}\) is said to be semi-simple if the corresponding Lie algebra \(\mathfrak{L}_{\mathrm{Lie}}\) is semi-simple. The authors define the Leibniz cohomology, \(HL^*(\mathfrak{L};\,M)\), of the left Leibniz algebra \(\mathfrak{L}\) with coefficients in an \(\mathfrak{L}\)-bimodule \(M\) by using the mirror image of the coboundary map in [loc. cit.] for the Leibniz cohomology of a right Leibniz algebra \(\mathfrak{L}_R\) with coefficients in a representation of \(\mathfrak{L}_R\). The two theories are isomorphic. For reference, a bimodule \(M\) over a left Leibniz algebra \(\mathfrak{L}\) is a \(k\)-vector space supporting a left and right action \[\mathfrak{L} \times M \to M, \ \ \ M \times \mathfrak{L} \to M \] that satisfy: \begin{align*} & (x \circ y ) \circ m = x \circ ( y \circ m ) - y \circ ( x \circ m) \\ & (x \circ m ) \circ y = x \circ ( m \circ y) - m \circ ( x \circ y) \\ & (m \circ x ) \circ y = m \circ ( x \circ y) - x \circ ( m \circ y ), \ \ \ m \in M, \ \ x, \ y \in \mathfrak{L}. \end{align*} For an \(\mathfrak{L}\)-bimodule \(M\), let \[ M^{\mathfrak{L}} := \{m \in M \mid m \circ x = 0, \ \forall x \in \mathfrak{L} \, \} \] be the subspace of right \(\mathfrak{L}\)-invariants. Proven in the paper is that for a finite-dimensional semi-simple left Leibniz algebra \(\mathfrak{L}\) over a characteristic zero field, and \(M\) an \(\mathfrak{L}\)-bimodule with \(M^{\mathfrak{L}} = 0\), then \(HL^n (\mathfrak{L} ; \, M) = 0\) for \(n \geq 0\). The proof is via a version of the Pirashvili (or Hochschild-Serre) spectral sequence [T. Pirashvili, Ann. Inst. Fourier 44, No. 2, 401–411 (1994; Zbl 0821.17023)] applied to the short exact sequence \[ 0 \longrightarrow\mathrm{Leib}(\mathfrak{L})\longrightarrow\mathfrak{L} \longrightarrow\mathfrak{L}_{\mathrm{Lie}}\longrightarrow 0. \] Also used in the proof is the vanishing theorem [P. Ntolo, C. R. Acad. Sci., Paris, Sér. I 318, No. 8, 707–710 (1994; Zbl 0797.17012)] for the Leibniz cohomology of a semi-simple Lie algebra in characteristic zero. Furthermore, if the hypothesis on \(M^{\mathfrak{L}}\) is dropped and \(M\) is a finite-dimensional \(\mathfrak{L}\)-bimodule, then \(HL^n (\mathfrak{L};\, M) = 0\) for \(n \geq 2\). In general the result does not hold over a field of prime characteristic with a counter-example given in the paper. For \(\mathbf{F}\) a field of characteristic zero, and \(\mathfrak{L}\) a finite-dimensional semi-simple left Leibniz algebra, \(HL^n (\mathfrak{L};\,\mathbf{F}) = 0\) for \(n \geq 1\). If \(\mathfrak{L}_{ad}\) denotes the adjoint representation of \(\mathfrak{L}\), then \(HL^n (\mathfrak{L}; \,\mathfrak{L}_{ad}) = 0\), \(n \geq 2\), confirming a conjecture of Adashev, Ladra and Omirov. The paper also contains calculations for the Leibniz cohomology of the two-dimensional non-abelian Lie algebra \(\mathfrak{A}\) on two symbols \(h\), \(e\) with \([h, \, e] = e = -[e, \, h]\). Coefficients are in a one-dimensional left \(\mathfrak{A}\)-module \(\mathbf{F}_{\lambda}= \langle 1 \rangle\) with \(h(1) = \lambda\) and \(e(1) = 0\).