1896 United States presidential election in Rhode Island
The 1896 United States presidential election in Rhode Island took place on November 3, 1896. Voters chose 4 representatives, or electors to the Electoral College, who voted for president and vice president.
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Elections in Rhode Island | ||||||
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Rhode Island 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 wide margin of 41.94%.
Bryan, running on a platform of free silver, appealed strongly to Western miners and farmers in the 1896 election, but had little appeal in Northeastern states like Rhode Island.
With 68.33% of the popular vote, Rhode Island would be McKinley's fourth strongest victory in terms of percentage in the popular vote after Vermont, neighboring Massachusetts and New Hampshire.[1]
Results
1896 United States presidential election in Rhode Island[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | William McKinley of Ohio | Garret Hobart of New Jersey | 37,437 | 68.33% | 4 | 100.00% | ||
Democratic | William Jennings Bryan of Nebraska | Arthur Sewall of Maine | 14,459 | 26.39% | 0 | 0.00% | ||
National Democratic | John McAuley Palmer of Illinois | Simon Bolivar Buckner of Kentucky | 1,166 | 2.13% | 0 | 0.00% | ||
Prohibition | Joshua Levering of Maryland | Hale Johnson of Illinois | 1,160 | 2.12% | 0 | 0.00% | ||
Socialist Labor | Charles Horatio Matchett of New York | Matthew Maguire of New Jersey | 558 | 1.02% | 0 | 0.00% | ||
N/A | Others | Others | 5 | 0.01% | 0 | 0.00% | ||
Total | 54,785 | 100.00% | 4 | 100.00% |
References
- "1896 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1896 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.