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1864 United States presidential election in Vermont

From Wikipedia, the free encyclopedia

1864 United States presidential election in Vermont

← 1860 November 8, 1864 1868 →
 
Abraham Lincoln November 1863.jpg
GeorgeMcClellan2.jpg
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois New Jersey
Running mate Andrew Johnson George H. Pendleton
Electoral vote 5 0
Popular vote 42,420 13,322
Percentage 76.10% 23.90%

President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

The 1864 United States presidential election in Vermont took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Vermont voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a wide margin of 52.20%.

With 76.10% of the popular vote, Lincoln's victory with in the state would be his second strongest victory in the country in terms of percentage in the popular vote after Kansas.[1]

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  • ✪ The American Presidential Election of 1860
  • ✪ The American Presidential Election of 2016
  • ✪ The American Presidential Election of 1876
  • ✪ The American Presidential Election of 1840
  • ✪ The American Presidential Election of 1908

Transcription

Results

1864 United States presidential election in Vermont[2]
Party Candidate Votes Percentage Electoral votes
National Union Abraham Lincoln 42,420 76.10% 5
Democratic George B. McClellan 13,322 23.90% 0
Totals 55,742 100.0% 26

References

  1. ^ "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1864 Presidential General Election Results - Vermont".
This page was last edited on 1 March 2020, at 21:05
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