1860 United States presidential election in Vermont

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1860 United States presidential election in Vermont

← 1856 November 2, 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 Vespasian Johnson
Electoral vote 5 0
Popular vote 33,808 8,649
Percentage 75.86% 19.41%

County Results

President before election

James Buchanan

Elected President

Abraham Lincoln

The 1860 United States presidential election in Vermont took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose five electors of the Electoral College, who voted for president and vice president.

Vermont was won by Republican candidate Abraham Lincoln, who won the state by a 56.45% margin.

With 75.86 percent of the popular vote, Lincoln's victory with in the state would be his strongest victory in terms of percentage in the popular vote.[1]

Northern Democratic presidential candidate Stephen A. Douglas was born in Vermont, more specifically in the town of Brandon.


1860 United States presidential election in Vermont[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Abraham Lincoln of Illinois Hannibal Hamlin of Maine 33,808 75.86% 5 100%
Democratic Stephen A. Douglas of Illinois Herschel Vespasian Johnson of Georgia 8,649 19.41% 0 0.00%
Southern Democratic John C. Breckinridge of Kentucky Joseph Lane of Oregon 1,866 4.19% 0 0.00%
Constitutional Union John Bell of Tennessee Edward Everett of Massachusetts 217 0.49% 0 0.00%
N/A Others Others 26 0.06% 0 0.00%
Total 44,566 100% 5 100%


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