1860 United States presidential election in Rhode Island

James Buchanan Democratic Abraham Lincoln Republican The 1860 United States presidential election in Rhode Island took place on November 2, 1860, as part of the 1860 United States presidential election.

Republican John C. Breckinridge/ John Bell Fusion 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.

[3] 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.

This Rhode Island elections-related article is a stub.