Stan Wawrinka vs Alexander Zverev AO 2020 Preview and Prediction

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Stan Wawrinka Australian Open

Alexander Zverev‘s slam breakthrough may finally have come as he looks to book his place in a semi-final for the first time on Wednesday. He will play Stan Wawrinka in what is sure to be a heavy hitting affair. The German has won both their matches, the most recent coming in 2017.

Zverev has been one of the most impressive players in the tournament and is now the only player to have not lost a set so far after four rounds. He had seen off some weaker opponents through his first three matches but stepped up to the task against an in-form Andrey Rublev in the fourth round, controlling the match in his 6-4 6-4 6-4 win. Zverev didn’t face a single break point in the match, meaning a solitary break would be enough in each set. He admits that he has been improving with match and has played himself into good form after a poor ATP Cup. “The second match wasn’t the best either, but I won. Then it went better and better. I hope it will continue being like that.” he said.

Wawrinka rolled back the years on Monday, picking up a surprise five set win over Daniil Medvedev. He had been struggling for consistent form since a run of injuries but showed he is more than capable of hanging with the best with this performance, much like his win over Novak Djokovic had done in New York last year. “I have been practising really hard,” Wawrinka said. “I’ve been feeling great on the court, on the practice court, good sessions, putting in a lot of work physically and tennis-wise. The week before I came here, I was really feeling at the top level.”

It will be difficult to chose between the two players after they come off such impressive but different wins over their Russian opponents. If Zverev can keep up the level he displayed on Monday, he may well just edge it but Wawrinka remains one of the top big match players on tour and could make it nervy for an opponent not used to the pressure the longer the match goes.

Prediction – Alexander Zverev in four sets

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