1856 United States presidential election in Illinois explained

See main article: 1856 United States presidential election.

Election Name:1856 United States presidential election in Illinois
Country:Illinois
Flag Year:1856
Type:presidential
Ongoing:no
Previous Election:1852 United States presidential election in Illinois
Previous Year:1852
Next Election:1860 United States presidential election in Illinois
Next Year:1860
Election Date:November 4, 1856
Image1:James Buchanan (cropped).jpg
Nominee1:James Buchanan
Party1:Democratic Party (United States)
Home State1:Pennsylvania
Running Mate1:John C. Breckinridge
Electoral Vote1:11
Popular Vote1:105,528
Percentage1:44.09%
Nominee2:John C. Frémont
Party2:Republican Party (United States)
Home State2:California
Running Mate2:William L. Dayton
Electoral Vote2:0
Popular Vote2:96,275
Percentage2:40.23%
Image3:Fillmore (cropped).jpg
Nominee3:Millard Fillmore
Party3:Know Nothing
Home State3:New York
Running Mate3:Andrew J. Donelson
Electoral Vote3:0
Popular Vote3:37,531
Percentage3:15.68%
Map Size:350px
President
Before Election:Franklin Pierce
Before Party:Democratic Party (United States)
After Election:James Buchanan
After Party:Democratic Party (United States)

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 11 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 narrow margin of 3.86%.

This would be the final time a Democratic presidential candidate would win Illinois until Grover Cleveland won it in 1892.

See also