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.
Rhode Island was won by Republican candidate Abraham Lincoln, who won by a margin of 22.74%.
Results 1860 United States presidential election in Rhode IslandPartyCandidateVotesPercentageElectoral votes Republican Abraham Lincoln 12,244 61.37% 4 Fusion Stephen A. Douglas / John C. Breckinridge / John Bell 7,707 38.63% 0 Totals 19,951 100.00% 4
Results by county +1860 United States Presidential Election in Rhode Island (by county)CountyAbraham LincolnRepublicanStephen A. Douglas/John C. Breckinridge/John BellFusionTotal votescast#%#%Bristol 66759.08%46240.92%1,129Kent 1,24665.48%65734.52%1,903Newport 1,61064.68%87935.32%2,489Providence 7,20259.63%4,87540.37%12,077Washington 1,51964.56%83435.44%2,353Total12,24461.37%7,70738.63%19,951
Analysis 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, neighboring Massachusetts and Maine.
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 not only 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.
See also United States presidential elections in Rhode Island
References