Edward Zakrzewski, convicted of the murder of his wife and children in 1994, is scheduled to be executed on July 31 in Florida. This is the ninth execution to take place this year.
Highlights
- Edward Zakrzewski was convicted of triple murder in 1994.
- The execution will take place on July 31 in Florida.
- This is the ninth execution this year in this state.
Edward J. Zakrzewski II was sentenced to death for the brutal murder of his wife, Sylvia, and their two children in June 1994 in Okaloosa County, Florida. The tragedy occurred after the man’s wife decided to file for divorce. Zakrzewski pleaded guilty. The court handed down three death sentences in his case for each crime he committed.
Florida Governor Ron DeSantis signed an order to execute the death penalty on Tuesday. The execution of Edward Zakrzewski was scheduled for the 31st day of July. This is the ninth execution this year in the state, which shows that the local authorities are consistently carrying out the death sentences imposed by the courts.
As the AP website reminds, court documents show that Zakrzewski killed his wife with a crowbar and a machete. He was also supposed to strangle her. The children, seven-year-old son Edward and five-year-old daughter Anna, died from blows inflicted on them by the perpetrator with a machete. After the crime, the man went into hiding for some time, but eventually turned himself in to the police. He was prompted to do so by the broadcast of the “Unsolved Mysteries” program, in which his case was presented.
