1860 United States presidential election in Minnesota

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.

1860 United States presidential election in Minnesota

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

County Results

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

Minnesota voted in its first ever United States presidential election, having been admitted as the 32nd state on May 11, 1858. The state 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 (DIllinois), running with 41st Governor of Georgia Herschel V. Johnson, with 43.97% of the popular vote.

With 63.53% 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]

Results

1860 United States presidential election in Minnesota[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%
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References

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