1856 United States presidential election in Rhode Island
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 County Results
Frémont 50-60% 60-70%
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| Elections in Rhode Island |
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The 1856 United States presidential election in Rhode Island took place on November 4, 1856, as part of the 1856 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the Republican candidate, John C. Frémont, over the Democratic candidate, James Buchanan, and the Know Nothing candidate, Millard Fillmore. Frémont won the state by a margin of 24.15%.
With 57.85% of the popular vote, Rhode Island proved to be Frémont's fourth strongest state in the 1856 election after Vermont, Massachusetts and Maine.[1]
Results
[edit]| 1856 United States presidential election in Rhode Island[2] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | John C. Frémont of California | William L. Dayton of New Jersey | 11,467 | 57.85% | 4 | 100.00% | ||
| Democratic | James Buchanan of Pennsylvania | John C. Breckinridge of Kentucky | 6,680 | 33.70% | 0 | 0.00% | ||
| Know Nothing | Millard Fillmore of New York | Andrew Jackson Donelson of Tennessee | 1,675 | 8.45% | 0 | 0.00% | ||
| Total | 19,822 | 100.00% | 4 | 100.00% | ||||
See also
[edit]References
[edit]- ^ "1856 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1856 Presidential General Election Results - Rhode Island".
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