Felipe Escudero Donoso vs Carlos Eduardo Pollio
2012 · Result 1–0 · Queen's Pawn Game: Zukertort Variation (D02).
Turn this game into your next win
Replay Felipe Escudero Donoso vs Carlos Eduardo Pollio with deep analysis, save the moments that matter, fold the ideas into your own opening repertoire, and drill the positions until they're second nature. CipherChess turns the games you study into the results you get — free to start.
Start Free on CipherChessMore Games By These Players
Game details
- White
- Felipe Escudero Donoso (2126)
- Black
- Carlos Eduardo Pollio (1989)
- Result
- 1–0
- Year
- 2012
- Opening
- Queen's Pawn Game: Zukertort Variation (D02)
About this chess game
This chess game between Felipe Escudero Donoso (2126) and Carlos Eduardo Pollio (1989) was played in 2012 and finished 1–0. The opening was the Queen's Pawn Game: Zukertort Variation (D02). You can replay the full game move by move on the interactive board above, or open it on the CipherChess analysis board to study every move with the Stockfish engine.
Looking for more Felipe Escudero Donoso games or Carlos Eduardo Pollio games? This Felipe Escudero Donoso vs Carlos Eduardo Pollio encounter is one of millions of chess games indexed in the CipherChess mega database. Browse both players' full records, the openings they play most, and head-to-head results, then load any game onto the board to prepare your own lines against the Queen's Pawn Game: Zukertort Variation.
Frequently asked questions
Who won Felipe Escudero Donoso vs Carlos Eduardo Pollio?
Felipe Escudero Donoso vs Carlos Eduardo Pollio (2012) finished 1–0, a win for Felipe Escudero Donoso.
What opening was played in Felipe Escudero Donoso vs Carlos Eduardo Pollio?
The game opened with the Queen's Pawn Game: Zukertort Variation (ECO D02).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Felipe Escudero Donoso vs Carlos Eduardo Pollio, or open it on the CipherChess analysis board to review it with the Stockfish engine.