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 four electors of the Electoral College, who voted for president and vice president.

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1860 United States presidential election in Rhode Island

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Nominee Abraham Lincoln Stephen Douglas (Dem) John C. Breckinridge (SD) John Bell (CU) Party Republican Fusion Alliance Democratic Southern Democratic Constitutional Union Home state Illinois Illinois (Douglas) Kentucky (Breckinridge) Tennessee (Bell) Running mate Hannibal Hamlin Herschel V. Johnson (Dem) Joseph Lane (SD) Edward Everett (CU) Electoral vote 4 0 Popular vote 12,244 7,707 Percentage 61.37% 38.63%

County Results Lincoln   50-60%   60-70%

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

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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 was supported just by the Northern Democrats but also supporters of Southern Democrats and Constitutional Unionists. However, unlike in the other three states the electors on the Democratic ticket in Rhode Island were pledged solely to Douglas, therefore some sources credit the Democratic popular vote to the Northern Democratic nominee.

• United States presidential elections in Rhode Island