1860 United States presidential election in Vermont

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1860 United States presidential election in Vermont
Flag of Vermont (1837-1923).svg
  1856 November 2, 1860 1864  
  Abraham Lincoln by Alexander Hesler, 1860-restored (cropped).png Senator Stephen A. Douglas (edited).png
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote50
Popular vote33,8088,649
Percentage75.86%19.41%

Vermont Presidential Election Results 1860.svg
County Results
Lincoln
  60-70%
  70-80%
  80-90%

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.

Contents

Vermont was won by Republican candidate Abraham Lincoln and his running mate Hannibal Hamlin They defeated Democratic candidate Stephen A. Douglas and his running mate Herschel V. Johnson. Lincoln won the state by a landslide margin of 56.45%.

With 75.86% of the popular vote, Vermont would be Lincoln's strongest victory in terms of percentage in the popular vote. [1]

Stephen A. Douglas was born in Brandon, Vermont.

Results

1860 United States presidential election in Vermont [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Republican Abraham Lincoln of Illinois Hannibal Hamlin of Maine 33,80875.86%5100%
Democratic Stephen Arnold Douglas of Illinois Herschel Vespasian Johnson of Georgia 8,64919.41%00.00%
Southern Democratic John Cabell Brekinridge of Kentucky Joseph Lane of Oregon 1,8664.19%00.00%
Constitutional Union John Bell of Tennessee Edward Everett of Massachusetts 2170.49%00.00%
N/AOthersOthers260.06%00.00%
Total44,566100%5100%

See also

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References

  1. "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. "1860 Presidential General Election Results - Vermont".