1896 United States presidential election in Delaware

Main article: 1896 United States presidential election

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

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

• United States presidential elections in Delaware

Notes