1896 United States presidential election in New Jersey
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| Elections in New Jersey |
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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
- ^ "1896 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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