1860 United States presidential election in Illinois

The 1860 United States presidential election in Illinois took place on November 6, 1860, as part of the 1860 United States presidential election. Illinois voters chose eleven representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1860 United States presidential election
1860 United States presidential election in Illinois
November 6, 1860
 
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote 11 0
Popular vote 172,171 160,215
Percentage 50.69% 47.17%
President before election

James Buchanan
Democratic

 Elected President

Abraham Lincoln
Republican

Elections in Illinois

Illinois was won by Illinois Representative Abraham Lincoln (R–Kentucky), running with Senator Hannibal Hamlin, with 50.69% of the popular vote, against Senator Stephen A. Douglas (D–Illinois), running with 41st Governor of Georgia Herschel V. Johnson, with 47.17% of the popular vote.

Results
1860 United States presidential election in Illinois[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Abraham Lincoln of Illinois Hannibal Hamlin of Maine 172,171 50.69% 11 100.00%
Democratic Stephen A. Douglas of Illinois Herschel Vespasian Johnson of Georgia 160,215 47.17% 0 0.00%
Constitutional Union John Bell of Tennessee Edward Everett of Massachusetts 4,914 1.45% 0 0.00%
Southern Democratic John C. Breckinridge of Kentucky Joseph Lane of Oregon 2,331 0.69% 0 0.00%
Liberty Gerrit Smith of New York Samuel McFarland of Pennsylvania 35 0.01% 0 0.00%
Total 339,666 100.00% 11 100.00%

References
  1. "1860 Presidential General Election Results - Illinois".
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