1860 United States presidential election in Delaware

The 1860 United States presidential election in Delaware took place on November 6, 1860, as part of the 1860 United States presidential election. Delaware voters chose 3 representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1860 United States presidential election

1860 United States presidential election in Delaware

November 6, 1860

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%

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

Delaware was won by the 14th Vice President of the United States John C. Breckinridge (SD–Kentucky), running with Senator Joseph Lane, with 45.54% of the popular vote, against Senator John Bell (CU–Tennessee), running with the Governor of Massachusetts Edward Everett, with 24.13% of the popular vote, Illinois Representative Abraham Lincoln (R–Kentucky), running with Senator Hannibal Hamlin, with 23.72% of the popular vote and the 15th Senator Stephen A. Douglas (D–Vermont), running with 41st Governor of Georgia Herschel V. Johnson, with 6.61% of the popular vote.

Results

1860 United States presidential election in Delaware[1]

Party		Candidate	Votes	%
	Southern Democratic	John C. Breckinridge	7,339	45.54%
	Constitutional Union	John Bell	3,888	24.13%
	Republican	Abraham Lincoln	3,822	23.72%
	Democratic	Stephen A. Douglas	1,066	6.61%
Total votes			16,115	100.00%

References

1. "1860 Presidential Election Results Delaware".