1860 United States presidential election in Vermont

Summary

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.

1860 United States presidential election in Vermont

← 1856 November 2, 1860 1864 →
 
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote 5 0
Popular vote 33,808 8,649
Percentage 75.86% 19.41%

County Results
Lincoln
  60-70%
  70-80%
  80-90%


President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

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 edit

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 Arnold Douglas of Illinois Herschel Vespasian Johnson of Georgia 8,649 19.41% 0 0.00%
Southern Democratic John Cabell Brekinridge 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%

See also edit

References edit

  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".