Pittsburgh Steelers quarterback Ben Roethlisberger has been named AFC Offensive Player of the Week for his stellar performance against the Cincinnati Bengals last Sunday.
Roethlisberger completed 27-of-46 passes for 333 yards and four touchdowns with no interceptions, leading the Steelers to the 36-10 win.
Adding to his impressive outing, Roethlisberger did not practice once leading up to the victory over Cincinnati after he was placed on the Reserve/COVID-19 list on Tuesday. He was able to participate in the team’s walkthrough once he was activated off the list on Saturday.
This is the 18th time Roethlisberger has been named AFC Offensive Player of the Week. He was last awarded the honor in Week 10 of 2018.
He is the fourth Steeler to be awarded Player of the Week honors this season. Rookie wide receiver Chase Claypool received it following his breakout four-touchdown performance versus Philadelphia. Outside linebacker T.J. Watt and defensive end Stephon Tuitt have both been named AFC Defensive Player of the Week in 2020.