A Diceware passphrase is constructed from a set of words chosen from a 7776-word database, using five dice throws to choose each word. The argument is that each word in the phrase adds about 12.92 bits of entropy since $\log_2(7776)≈12.92$. As Arnold Reinhold, the creator of Diceware, explains, the entropy comes from the size of the database, not from the way in which the elements are represented. So, in principle, it does not matter whether a 7-word passphrase is formed as 7 English words from a Diceware database, the original 7 five-dice throws (in effect 7 five-digit base six numbers with each digit increased by 1), or seven Chinese characters chosen from a database of 7776 Chinese characters. Each would have an entropy of about 90 bits.
Is it not the case, however, that using English words in place of the underlying dice-throw numbers reduces the entropy of the passphrase. The words in the standard Diceware database have an average length of 4.2 characters. This means that the average length of a 7-word passphrase would be about 30 characters. It is said that letters in English text have an entropy of about 1.3 bits each due to the redundancy of the language. This would imply that, analyzed as English text, the passphrase would only have an entropy of 38 bits. Arguably a Diceware passphrase is not typical English text, but presumably one could do a similar analysis to determine the redundancy of text drawn from the Diceware database, but, if my argument is correct, each letter in the passphrase would have to have at least 3 bits of entropy in order not to downgrade the entropy of the underlying Diceware elements.
Reinhold himself glosses over this. He says:
But, you might ask, Diceware is made up mostly of English words. Doesn't that redundancy affect it at all? Well, it does. The seven word Diceware passphrase we talked about above would average about 30 letters in length (36 if you count the spaces between the words). If those letters were selected randomly, you would get 4.7 bits of strength per letter. That would work out to a lot more than the 90 bits Diceware claims for a seven word passphrase. The difference is the redundancy of English at work. However English redundancy does not affect the calculation that each Diceware word has 12.9 bits of randomness, which is based entirely how many different words there are in the Diceware list. You can rely on that number.
The redundancy of English does not affect the entropy of the Diceware passphrase, so long as you can rely on each letter having 4.7 bits of entropy, or indeed at least 3 bits. But I don't see what guarantee there is that this is the case.
Note that this question is not answered by Can an attacker exploit the nonuniform distribution of letters in a diceware password to improve a brute force search? since this question raises a different argument not considered in that question. The underlying issue is similar, but the question asked about it is different.