1860 United States presidential election in Indiana

The 1860 United States presidential election in Indiana took place on November 6, 1860, as part of the 1860 United States presidential election. Indiana voters chose 13 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 Indiana
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 13 0
Popular vote 139,033 115,509
Percentage 51.09% 42.44%
President before election

James Buchanan
Democratic

 Elected President

Abraham Lincoln
Republican

Elections in Indiana
Federal government
U.S. President
U.S. Senate
U.S. House of Representatives
State government
State elections
Gubernatorial elections
State Senate elections
State House of Representatives elections
Evansville
Mayoral Elections
Fort Wayne
Mayoral Elections
Gary
Mayoral Elections
Indianapolis
Mayoral Elections
Before Unigov
Unigov-era
City-County Council Elections
South Bend
Mayoral Elections

Indiana was won by Illinois Representative Abraham Lincoln (R–Kentucky), running with Senator Hannibal Hamlin, with 51.09% of the popular vote, against Senator Stephen A. Douglas (D–Vermont), running with 41st Governor of Georgia Herschel V. Johnson, with 42.44% of the popular vote.

Results
1860 United States presidential election in Indiana[1]
Party Candidate Votes %
Republican Abraham Lincoln 139,033 51.09%
Democratic Stephen A. Douglas 115,509 42.44%
Southern Democratic John C. Breckinridge 12,295 4.52%
Constitutional Union John Bell 5,306 1.95%
Total votes 272,143 100%

See also

References
  1. "1860 Presidential Election Results Indiana".
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