1896 United States presidential election in New Jersey

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United States presidential election in New Jersey, 1896

← 1892 November 3, 1896 1900 →
  William McKinley by Courtney Art Studio, 1896.jpg WilliamJBryan1902.png
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 10 0
Popular vote 221,535 133,695
Percentage 59.68% 36.02%

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in New Jersey took place on November 3, 1896. Voters chose 10 representatives, or electors to the Electoral College, who voted for president and vice president.

New Jersey voted for the Republican nominee, former governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. McKinley won the state by a margin of 23.66%, making him the first Republican presidential candidate since Ulysses S. Grant in 1872 to carry the state.

Results

United States presidential election in New Jersey, 1896[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican William McKinley of Ohio Garret Hobart of New Jersey 221,535 59.68% 10 100.00%
Democratic William Jennings Bryan of Nebraska Arthur Sewall of Maine 133,695 36.02% 0 0.00%
National Democratic John McAuley Palmer of Illinois Simon Bolivar Buckner of Kentucky 6,378 1.72% 0 0.00%
Prohibition Charles Eugene Bentley of Nebraska James Haywood Southgate of North Carolina 5,617 1.51% 0 0.00%
Socialist Labor Charles Horatio Matchett of New York Matthew Maguire of New Jersey 3,986 1.07% 0 0.00%
Total 371,211 100.00% 10 100.00%

References

  1. ^ "1896 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.


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