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1860 United States presidential election in Pennsylvania

From Wikipedia, the free encyclopedia

1860 United States presidential election in Pennsylvania

← 1856 November 6, 1860 1864 →
Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg
Nominee Abraham Lincoln Democratic electors
Party Republican Fusion
Home state Illinois Kentucky / Illinois
Running mate Hannibal Hamlin Joseph Lane /
Herschel Johnson
Electoral vote 27 0
Popular vote 268,030 178,871
Percentage 56.26% 37.54%

President before election

James Buchanan

Elected President

Abraham Lincoln

The 1860 United States presidential election in Pennsylvania took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose 27 representatives, or electors to the Electoral College, who voted for president and vice president.

The Democratic Party chose its slate of electors before the National Convention in Charleston, South Carolina. Since this was decided before the party split, both Douglas supporters and Breckinridge supporters claimed the right for their man to be considered the party candidate and the support of the electoral slate. Eventually, the state party worked out an agreement: if either candidate could win the national election with Pennsylvania's electoral vote, then all her electoral votes would go to that candidate. Of the 27 electoral candidates, 15 were Breckinridge supporters; the remaining 12 were for Douglas. This was often referred to as the Reading electoral slate, because it was in that city that the state party chose it.[1]

Not all of the Douglas supporters agreed to the Reading slate deal and established a separate Douglas-only ticket. This slate comprised the 12 Douglas electoral candidates on the Reading ticket, and 15 additional Douglas supporters. This ticket was usually referred to as the Straight Douglas ticket. Thus 12 electoral candidates appeared on 2 tickets, Reading and Straight Douglas.[1]

Pennsylvania voted for the Republican candidate, Abraham Lincoln, over the fusion ticket. Lincoln won Pennsylvania by a margin of 18.72%.

Pennsylvania in the election was one of the four states that had a fusion ticket for the Democratic Party. The other three states were New Jersey, New York and Rhode Island.

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  • ✪ The American Presidential Election of 1904
  • ✪ The American Presidential Election of 1852
  • ✪ The American Presidential Election of 1916
  • ✪ The American Presidential Election of 2012
  • ✪ The American Presidential Election of 1880



United States presidential election in Pennsylvania, 1860[2]
Party Candidate Votes Percentage Electoral votes
Republican Abraham Lincoln 268,030 56.26% 27
Fusion John C. Breckinridge / Stephen A. Douglas 178,871 37.54% 0
Northern Democratic Stephen A. Douglas 16,765 3.52% 0
Constitutional Union John Bell 12,776 2.68% 0
Totals 476,442 100.0% 27


  1. ^ a b Dubin, Michael J., United States Presidential Elections, 1788–1860: The Official Results by County and State, McFarland & Company, 2002, p. 188
  2. ^ "1860 Presidential General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved 3 August 2012.

This page was last edited on 17 May 2019, at 20:10
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