1896 United States presidential election in Delaware

The 1896 United States presidential election in Delaware took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.

1896 United States presidential election in Delaware

← 1892 November 3, 1896 1900 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 3 0
Popular vote 20,450 16,574
Percentage 53.18% 43.10%

County Results
McKinley
  50-60%


President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

Delaware was won by the Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey. They defeated the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan and his running mate Arthur Sewall. McKinley won the state by a margin of 10.08%.

As a result of his win in Delaware, McKinley became the first Republican presidential candidate since Ulysses S. Grant in 1872 to win the state.

Bryan would lose Delaware to McKinley again four years later and would later lose the state again in 1908 to William Howard Taft.

Results

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1896 United States presidential election in Delaware[1]
Party Candidate Votes Percentage Electoral votes
Republican William McKinley 20,450 53.18% 3
Democratic William Jennings Bryan 16,574 43.10% 0
National Democratic John M. Palmer 966 2.51% 0
Prohibition Joshua Levering 466 1.21% 0
Totals 38,456 100.00% 3
Voter turnout

See also

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Notes

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References

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  1. ^ Dave Leip's U.S. Election Atlas; Presidential General Election Results – Delaware