1860 United States presidential election in Rhode Island

The 1860 United States presidential election in Rhode Island took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose 4 electors of the Electoral College, who voted for president and vice president.

1860 United States presidential election in Rhode Island

← 1856 November 2, 1860 1864 →
  Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg DemocraticLogo.svg
Nominee Abraham Lincoln John C. Breckinridge /
Stephen A. Douglas
Party Republican Fusion
Home state Illinois Kentucky / Illinois
Running mate Hannibal Hamlin Joseph Lane /
Herschel Johnson
Electoral vote 4 0
Popular vote 12,244 7,707
Percentage 61.37% 38.63%

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

Rhode Island was won by Republican candidate Abraham Lincoln, who won by a margin of 22.74%.

With 61.37% of the popular vote, Rhode Island would prove to be Lincoln's fifth strongest state in terms of popular vote percentage in the 1860 election after Vermont, Minnesota, Massachusetts and Maine.[1]

Like New Jersey, New York and Pennsylvania, Rhode Island was one of the four states that had a fusion ticket for the Democrats, which consisted of not just the Northern Democrats, but of supporters of Southern Democrats and Constitutional Unionists as well.

ResultsEdit

1860 United States presidential election in Rhode Island[2]
Party Candidate Votes Percentage Electoral votes
Republican Abraham Lincoln 12,244 61.37% 4
Fusion John C. Breckinridge / Stephen A. Douglas 7,707 38.63% 0
Totals 19,951 100.0% 4


ReferencesEdit

  1. ^ "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1860 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 3 August 2012.