1860 United States presidential election in Minnesota

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United States presidential election in Minnesota, 1860

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 4 0
Popular vote 22,069 11,920
Percentage 63.53% 34.31%

Minnesota President Election Results by County, 1860.svg
  >= 40%
  >= 50%
  >= 60%
  >= 70%
  >= 80%
  >= 90%


  >= 50%
  >= 60%

Unknown/No Vote:


President before election

James Buchanan

Elected President

Abraham Lincoln

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

Minnesota was won by Illinois Representative Abraham Lincoln (Republican Party (United States)), running with Senator Hannibal Hamlin, with 57.23% of the popular vote, against Senator Stephen A. Douglas (DVermont), running with 41st Governor of Georgia Herschel V. Johnson, with 43.97% of the popular vote.

With 63.53 percent of the popular vote, Lincoln's victory within the state would be his second strongest victory in terms of percentage in the popular vote in the 1860 election after Vermont.[1]


United States presidential election in Minnesota, 1860[2]
Party Candidate Votes %
Republican Abraham Lincoln 22,069 63.53%
Democratic Stephen A. Douglas 11,920 34.31%
Southern Democratic John C. Breckinridge 748 2.15%
Constitutional Union John Bell 50 0.14%
Total votes 34,787 100%


  1. ^ "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1860 Presidential Election Results Minnesota".