Billable / Specific Code
ICD-10-CM S81.042 is the diagnosis code for Puncture wound with foreign body, left knee. This code falls under the section "Injuries to the knee and lower leg" within Chapter 19 — Injury, poisoning and certain other consequences of external causes (S00-T88). It is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. Medical coders and healthcare providers use this code to document and classify diagnoses in electronic health records, insurance claims, and clinical databases.
Always refer to the official ICD-10-CM Tabular List for complete coding guidelines. Ensure documentation supports the specificity of the code selected. When in doubt, consult a certified medical coder or the latest CMS guidelines.
Understanding where S81.042 sits in the ICD-10-CM classification helps ensure proper coding:
Yes, S81.042 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes.
S81.042 is the ICD-10-CM diagnosis code for "Puncture wound with foreign body, left knee". It is used by healthcare providers to classify and document this condition in medical records and insurance claims.
The parent code of S81.042 is S81.04 ("Puncture wound with foreign body of knee"). S81.042 provides a more specific classification within this category.
S81.042 is located in Section S80-S89 — "Injuries to the knee and lower leg" within Chapter 19 of the ICD-10-CM Tabular List.
Use S81.042 when the patients documented diagnosis matches "Puncture wound with foreign body, left knee" and the clinical documentation supports this level of specificity. Always verify with the latest ICD-10-CM guidelines and payer requirements.
The ICD-10-CM code for Puncture wound with foreign body, left knee is S81.042.
Yes, S81.042 can be used as a primary diagnosis code since it is billable and specific.
S81.042 is in Chapter 19 of the ICD-10-CM Tabular List.
Yes, S81.042 is a valid ICD-10-CM code for the 2026 fiscal year, subject to official CMS updates.