1860 United States presidential election in Vermont information
Election in Vermont
Main article: 1860 United States presidential election
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
Elections in Vermont
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U.S. House elections
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2nd sp
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1800
1802/03
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1st sp
1810
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AL sp
1826/27
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1832
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1836
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1936
1938
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AL sp
1974
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v
t
e
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 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.
^"1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
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