Using a one-time pad with a block cipher The one-time pad is often described with a simple XOR cipher. 

This is nice for showing the theoretical strengths of the one-time pad, but not so great in real-world scenarios where key reuse is a danger.

So, is it feasible to use an n-bit block cipher with an n-bit key to replace the XOR operation (which itself is a 1-bit block cipher with a 1-bit key)?

So the encryption operation becomes

for each pBlock is n bits of plaintext
     take kBlock as the corresponding n bits of the key
     encrypt pBlock with kBlock as key using the block cipher and the result is cBlock
     append cBlock to the ciphertext


Decryption is the same.

Am I correct that this will make it more resistant (but not fully) to key reuse and known plaintext attacks than using the simple XOR? Or did I add a big vulnerability into the scheme?
Well, no.

What you are describing is ECB mode of operation. ECB mode of operation is insecure as repeats of plaintext blocks leads to repeats of ciphertext blocks. This is a leak of information so it breaks confidentiality. So even if you would leak less information because of key reuse, you would still leak information if it happens.

Furthermore it is possible to find out if a key is (likely) correct by validating a ciphertext block against partial information within plaintext blocks. To that breaks the perfect secrecy provided by a One Time Pad.



I'll try and explain that last section further.

Note that there aren't $2^n$ possible permutations, there are $2^n!$ possible permutations. To be perfectly secure you should allow each one of them (with the same probability, no less).

That would mean that for any sensible $n$ the key size would grow out of proportion. Furthermore, you would have to have a cipher that would map all those keys to exactly one permutation, something that is not true for usual block ciphers.

So even if you have a key of n bits with m &lt; n bits of known plaintext then it is likely that you can tell with a probability higher than $0.5$ if it is the correct one or not, breaking perfect secrecy for the remaining n - m bits.