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1856 United States presidential election in Illinois

United States presidential election in Illinois, 1856

← 1852 November 4, 1856 1860 →
  Hon. James Buchanan - NARA - 528318-crop.jpg John Charles Fremont crop.jpg Millard Fillmore by Brady Studio 1855-65-crop.jpg
Nominee James Buchanan John C. Frémont Millard Fillmore
Party Democratic Republican Know Nothing
Home state Pennsylvania California New York
Running mate John C. Breckinridge William L. Dayton Andrew Jackson Donelson
Electoral vote 11 0 0
Popular vote 105,528 96,275 37,531
Percentage 44.09% 40.23% 15.68%

President before election

Franklin Pierce

Elected President

James Buchanan

The 1856 United States presidential election in Illinois took place on November 4, 1856, as part of the 1856 United States presidential election. Voters chose eleven representatives, or electors to the Electoral College, who voted for president and vice president.

Illinois voted for the Democratic candidate, James Buchanan, over Republican candidate John C. Frémont and American Party candidate Millard Fillmore. Buchanan won Illinois by a margin of 3.86%.


United States presidential election in Illinois, 1856[1][2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic James Buchanan of Pennsylvania John C. Breckinridge of Kentucky 105,528 44.09% 11 100.00%
Republican John C. Frémont of California William L. Dayton of New Jersey 96,275 40.23% 0 0.00%
Know Nothing Millard Fillmore of New York Andrew Jackson Donelson of Tennessee 37,531 15.68% 0 0.00%
Total 239,334 100.00% 11 100.00%


  1. ^ "1856 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 3 December 2017.
  2. ^ "1856 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved 3 December 2017.