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%.As neither John C. Breckinridge or John Bell were on the ballot in Rhode Island, most of their supporters threw their support to Douglas.

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dbo:abstract
  • 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%.As neither John C. Breckinridge or John Bell were on the ballot in Rhode Island, most of their supporters threw their support to Douglas. (en)
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  • United States presidential election in Rhode Island, 1856 (en)
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  • 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%.As neither John C. Breckinridge or John Bell were on the ballot in Rhode Island, most of their supporters threw their support to Douglas. (en)
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  • United States presidential election in Rhode Island, 1860 (en)
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