After putting together a shimmering performance that saw him complete 29-of-35 passes for 300 yards and three touchdowns in a win over the Steelers, Russell Wilson has been named the Offensive Player of the Week in the NFC. Wilson was joined by Buccaneers EDGE Shaquil Barrett (DPOTW) and Bears kicker Eddy Pineiro (STPOTW) as the winners in the conference for Week 2.
Wilson’s showing in Pittsburgh, which also included 22 yards on the ground, was as in-control as the signal-caller has ever looked. The game, particularly in the second half, was entirely dictated by him. The award for Week 2 is as deserved as his performance was enjoyable to watch as a viewer.
This is Wilson’s eighth time winning the award, which is the most in Seahawks franchise history—one more than former MVP Shaun Alexander. Wilson’s last player of the week award came in Week 13 of 2017, his second of two POTW awards in that season.
Wilson is the first in Seattle to be awarded a player of the week this season.