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1860 United States presidential election in Indiana

1860 United States presidential election in Indiana

← 1856 November 6, 1860 1864 →
  Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg BradyHandy-StephenADouglas restored.jpg
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

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 thirteen representatives, or electors, to the Electoral College, who voted for president and vice president.

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

Results [ edit ]

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%

References [ edit ]

  1. ^ "1860 Presidential Election Results Indiana".
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