Werder Bremen vs. Wolfsburg preview: The company club go for 7 in a row
By Jonathan Dicks @Jonathan_Dicks
In a battle between the Bundesliga's "green" teams, Wolfsburg travel to Bremen looking to continue their impressive winning run.
|Niko Kovac.||Photo: Sven Mandel / CC-BY-SA-4.0|
After the lengthy winter break, Werder have started slowly with a horrific 7-1 loss to Köln and a 2-1 defeat to Union Berlin at home. They’ve dropped to 11th in the table and now have the hottest team in the Bundesliga coming to town on Saturday.
Werder desperately need the strike pair of Niclas Füllkrug and Marvin Ducksch to return to the players we saw in the first half of the season. But still, with 21 points recorded from 17 games, Ole Werner has done a terrific job in giving this team a fantastic chance to stay in the league.
Werner commented, "Of course the mood isn't so good after defeats, but we are still very optimistic. We knew that moments like these could come for a promoted team. We are looking ahead."
Romano Schmid and Felix Agu are still unavailable for Matchday 18. Leonardo Bittencourt missed Werder’s final training session and is questionable.
Wolfsburg have won six Bundesliga games in a row and have rocketed up to 7th in the standings. No one could have predicted that Niko Kovac, who was about a game away from getting fired a couple months ago, could have turned things around in such eye-popping fashion.
And the company club aren’t just squeaking by teams. They slammed Freiburg 6-0 last weekend and then throttled Hertha 5-0 midweek. This is a scary squad in the form of their lives with a consistent defense and an endless amount of attacking options.
Kovac stated, “We still have 17 games to play and a lot can happen. That's why we are not reacting exuberantly now, but remain modest and above all, focused.”
Brothers Lukas and Felix Nmecha are unlikely to suit up on Saturday. Paulo Otavio will return to the squad after serving a suspension.
Wolfsburg have just signed left-back Nicolas Cozza from Montpellier HSC for €500,000.