1860 United States presidential election in Delaware information
Election in Delaware
Main article: 1860 United States presidential election
1860 United States presidential election in Delaware
← 1856
November 6, 1860
1864 →
Nominee
John C. Breckinridge
John Bell
Party
Southern Democratic
Constitutional Union
Home state
Kentucky
Tennessee
Running mate
Joseph Lane
Edward Everett
Electoral vote
3
0
Popular vote
7,339
3,888
Percentage
45.54%
24.13%
Nominee
Abraham Lincoln
Stephen A. Douglas
Party
Republican
Democratic
Home state
Illinois
Illinois
Running mate
Hannibal Hamlin
Herschel V. Johnson
Electoral vote
0
0
Popular vote
3,822
1,066
Percentage
23.72%
6.61%
County Results
Breckinridge
40-50%
50-60%
President before election
James Buchanan
Democratic
Elected President
Abraham Lincoln
Republican
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v
t
e
The 1860 United States presidential election in Delaware took place on November 6, 1860, as part of the 1860 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president.
Delaware was won by the Southern Democratic candidate 14th Vice President of the United States John C. Breckinridge of Kentucky and his running mate Senator Joseph Lane of Oregon. They defeated the Constitutional Union candidate Senator John Bell of Tennessee and his running mate Governor of Massachusetts Edward Everett, Republican candidate Illinois Representative Abraham Lincoln and his running mate Senator Hannibal Hamlin of Maine as well as Democratic candidate 15th Senator Stephen A. Douglas of Illinois and his running mate 41st Governor of Georgia Herschel V. Johnson. Breckinridge won the state by a margin of 21.41%.
The Republicans ran a slate of electors for Lincoln, but did not nominate any candidates for other federal and state offices. The party instead endorsed the candidates of the People's Party, an anti-Democratic coalition that supported the Republican presidential ticket.[1]
^Abbott 1986, p. 17.
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