Alabama freshman quarterback Keelon Russell captured the Shaun Alexander Award Player of the Week honors following a standout performance in the Crimson Tide’s 50‑36 triumph over Florida State on Saturday.
Stat‑filled showing‑off
Russell completed 23 of 30 pass attempts for 320 yards and three touchdowns, all delivered in the final nine minutes of the second quarter. His touchdown passes traveled 21, 46 and 16 yards to three different receivers, demonstrating both accuracy and big‑play ability.
In addition to his aerial success, Russell proved a dual‑threat with his legs, rushing three times for 56 yards and a rushing touchdown. The combination of passing precision and explosive runs helped Alabama surge ahead and maintain control throughout the game.
Shaun Alexander Award significance
The Shaun Alexander Award recognizes the most outstanding freshman player in college football each week. Russell, who redshirted at the end of last season after appearing in just two games, has quickly become a key contributor to Alabama’s offense. His performance helped the Crimson Tide start the 2026 season 3‑0 and climb back into the top ten of the AP Poll after opening the year at No. 13.
Looking ahead
Alabama, currently ranked No. 8, will host South Carolina on Saturday, September 26, at Bryant‑Denny Stadium. Kickoff is set for 6:00 p.m. on ESPN. Fans can expect Russell to continue leading the offense as the Tide pursue another championship‑contending season.
Wyatt Fulton, the Tide’s 100.9 DME and Brand Manager, provided the release and can be followed on X (formerly Twitter) at @FultonW_ for additional Crimson Tide coverage.
Original reporting: The Tuscaloosa Thread — read the source article.
