1864 United States presidential election in Delaware

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1864 United States presidential election in Delaware
Flag of Delaware.svg
  1860 November 8, 1864 1868  
  GeorgeMcClellan2 (cropped).jpg Abraham Lincoln November 1863.jpg
Nominee George B. McClellan Abraham Lincoln
Party Democratic National Union
Home state New Jersey Illinois
Running mate George H. Pendleton Andrew Johnson
Electoral vote30
Popular vote8,7678,155
Percentage51.81%48.19%

Delaware Presidential Election Results 1864.svg
County Results

President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

The 1864 United States presidential election in Delaware took place on November 8, 1864, as part of the 1864 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president. [1]

Contents

Delaware was won by the Democratic nominee, 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton. They defeated the National Union nominee, incumbent President Abraham Lincoln of Illinois and his running mate Senator and Military Governor of Tennessee Andrew Johnson. [1] McClellan won the state by a margin of 3.62%.

With 51.81% of the popular vote, Delaware would prove to be McClellan's third strongest state after Kentucky and New Jersey, his only two other winning states. [2]

Results

1864 United States presidential election in Delaware [1]
PartyCandidateVotes%
Democratic George B. McClellan 8,767 51.81%
National Union Abraham Lincoln (incumbent)8,15548.19%
Total votes16,922 100.00%

See also

References

  1. 1 2 3 "1864 Presidential Election Results Delaware".
  2. "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.