1896 United States presidential election in Nebraska

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United States presidential election in Nebraska, 1896

← 1892 November 3, 1896 1900 →
  WilliamJBryan1902.png William McKinley by Courtney Art Studio, 1896.jpg
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

Elected President

William McKinley

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. Nebraska voters chose eight electors to the Electoral College, which selected the president and vice president.

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


United States presidential election in Nebraska, 1896[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


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