RSA public key can decrypt the ciphertext it encrypted?
The prime 2 is special, and too small.

You happened to choose an $e$ which is equal to $d,$ as I explain below.
For a proper choice of large primes, it is extremely unlikely that $d=e,$ but if it is, one can simply choose another $e$ and compute the corresponding $d.$

Also, a correction, you public key is $(n,e)=(14,5)$. It is your ciphertext which is $c=m^5~mod~14.$ For this tiny example it happens that $e=5$ satisfies $$e^2=1~(mod~\phi(14))$$ so $d=e$, thus your public and private exponents are the same, not a good thing.

Also you say your private key is $m^{11}~mod~14$, which I interpret to mean your private key $(d,p,q)$ has $d=11.$ This is a valid $d$ since $5\times 11=55=1~mod~\phi(14).$ The correct terminology is $$m=c^d~(mod ~n)$$since the letter $m$ is reserved for the message.