1896 United States presidential election in Nebraska

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

Main article: 1896 United States presidential election

1896 United States presidential election in Nebraska

November 3, 1896

Nominee William J. Bryan William McKinley Party Democratic Republican Home state Nebraska Ohio Running mate Arthur Sewall Garret Hobart Electoral vote 8 0 Popular vote 115,007 103,064 Percentage 51.53% 46.18%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

Nebraska was won by the Democratic nominees, former U.S. Representative and Nebraska native William Jennings Bryan and his running mate Arthur Sewall of Maine. Four electors cast their Vice Presidential ballots for Thomas E. Watson.

As a result of his win, Bryan became the first Democratic presidential candidate to win Nebraska. Bryan would later lose the state to William McKinley in 1900 but would later win it again against William Howard Taft in 1908.

Results

1896 United States presidential election in Nebraska[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	William Jennings Bryan	115,007	51.53%	8
	Republican	William McKinley	103,064	46.18%	0
	National Democratic	John M. Palmer	2,885	1.29%	0
	Prohibition	Joshua Levering	1,243	0.56%	0
	National Prohibition	Charles Bentley	797	0.36%	0
	Socialist Labor	Charles Matchett	186	0.08%	0
Totals			223,182	100.00%	8
Voter turnout					—

References

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

Notes