Chelsea fans’ preferred lineup against Tottenham: Lukaku to get his chance; Kanté rested

It’s back to football as the first leg of the Carabao Cup semifinal sees Chelsea host Antonio Conte’s Tottenham Hotspur at the Bridge. The Blues looked severely fatigued at the end of Sunday’s game, so some degree of rotation seems inevitable. Maybe the start of a redemption arc (and goals) by a certain someone could revitalize some tired legs?

THE PEOPLE’S CHOICE

Édouard Mendy has traveled to the Africa Cup of Nations and will be unavailable for the next month. This means that the WAGNH community reinstate Kepa Arrizabalaga, who now has a chance to stake a claim while Marcus Bettinelli (2%) plays second fiddle. Formation-wise, the 3-4-3 remains unchallenged ahead of the 3-5-2 (17%).

With both Trevoh Chalobah and Andreas Christensen questionable at best, Malang Sarr comes in to the back three alongside Thiago Silva and Antonio Rüdiger. The dire situation at right wing-back sees Marcos Alonso and César Azpilicueta start again, just as on Sunday. Youngsters Xavier Simons (18%) and Lewis Hall (10%) did receive respectable amounts of votes.

After a truly astonishing performance against Liverpool, Mateo Kovačić unsurprisingly finds himself in the first eleven once again, this time next to Jorginho. N’Golo Kanté (41%) get a rest, while Lewis Baker (10%), Ross Barkley (20%), Harvey Vale (14%) and a possibly fit Ruben Loftus-Cheek (5%) provide options from the bench.

Romelu Lukaku might not be the most popular person around Stamford Bridge right now, but the congested schedule doesn’t care about such things — and neither did the community, making him the third highest vote-getter after Azpilicueta and Rüdiger. Mason Mount and Hakim Ziyech flank the striker, with Christian Pulisic (41%) and Kai Havertz (32%) dropping out to join a returning Timo Werner (27%).

3-4-3 (75%)
Arrizabalaga (98%) | Rüdiger (84%), Silva (77%), Sarr (56%) | Alonso (64%), Jorginho (65%), Kovačić (59%), Azpilicueta (89%) | Mount (41%), Lukaku (83%), Ziyech (56%)