  Let $F= < a,b>$ be a rank two free group. A word $W(a,b)$ in $F$ is {\sl
primitive} if it, along with another group element, generates the group. It is
a {\sl palindrome} (with respect to $a$ and $b$) if it reads the same forwards
and backwards. It is known that in a rank two free group any primitive element
is conjugate either to a palindrome or to the product of two palindromes, but
known iteration schemes for all primitive words give only a representative for
the conjugacy class. Here we derive a new iteration scheme that gives either
the unique palindrome in the conjugacy class or expresses the word as a unique
product of two unique palindromes. We denote these words by $E_{p/q}$ where
$p/q$ is rational number expressed in lowest terms. We prove that $E_{p/q}$ is
a palindrome if $pq$ is even and the unique product of two unique palindromes
if $pq$ is odd. We prove that the pairs $(E_{p/q},E_{r/s})$ generate the group
when $|ps-rq|=1$. This improves the previously known result that held only for
$pq$ and $rs$ both even. The derivation of the enumeration scheme also gives a
new proof of the known results about primitives.
