1856 United States presidential election in Rhode Island

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United States presidential election in Rhode Island, 1856

← 1852 November 4, 1856 1860 →
  JCFrémont.jpg Hon. James Buchanan - NARA - 528318-crop.jpg Millard Fillmore by Brady Studio 1855-65-crop.jpg
Nominee John C. Frémont James Buchanan Millard Fillmore
Party Republican Democratic Know Nothing
Home state California Pennsylvania New York
Running mate William L. Dayton John C. Breckinridge Andrew J. Donelson
Electoral vote 4 0 0
Popular vote 11,467 6,680 1,675
Percentage 57.85% 33.70% 8.45%

President before election

Franklin Pierce

Elected President

James Buchanan

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]


United States presidential election in Rhode Island, 1856[2]
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% 8 100.00%
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%


  1. ^ "1856 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1856 Presidential General Election Results - Rhode Island".

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