1860 United States presidential election in Maryland

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United States presidential election in Maryland, 1860

← 1856 November 6, 1860 1864 →
  John C Breckinridge-04775-restored.jpg John-bell-brady-handy-cropped restored.jpg BradyHandy-StephenADouglas restored.jpg
Nominee John Breckenridge John Bell Stephen A. Douglas
Party Southern Democratic Constitutional Union Democratic
Home state Kentucky Tennessee Illinois
Running mate Joseph Lane Edward Everett Herschel V. Johnson
Electoral vote 8 0 0
Popular vote 42,482 41,760 5,966
Percentage 45.93% 45.14% 6.45%

President before election

James Buchanan

Elected President

Abraham Lincoln

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

Maryland was won by the 14th Vice President of the United States John Breckenridge (SDKentucky), running with Senator Joseph Lane, with 45.93% of the popular vote, against Senator John Bell (CUTennessee), running with the 15th Governor of Massachusetts Edward Everett, with 45.14% of the popular vote and Senator Stephen A. Douglas (DVermont), running with 41st Governor of Georgia Herschel V. Johnson, with 6.45% of the popular vote.

Despite coming in a distant fourth place with 2,294 votes Abraham Lincoln did receive over 2,000 more votes than John C. Frémont received in 1856 and would later win the state in 1864 with 55% of the vote.


United States presidential election in Maryland, 1860[1]
Party Candidate Votes %
Southern Democratic John C. Breckinridge 42,482 45.93%
Constitutional Union John Bell 41,760 45.14%
Democratic Stephen A. Douglas 5,966 6.45%
Republican Abraham Lincoln 2,294 2.48%
Total votes 92,502 100%


  1. ^ "1860 Presidential Election Results Maryland".