The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.Rhode Island voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%.

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  • The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.Rhode Island voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%. (en)
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  • National Union Party (en)
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  • Republican Party (en)
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  • Rhode Island (en)
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  • 1864-11-08 (xsd:date)
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  • United States presidential election in Rhode Island, 1864 (en)
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  • United States presidential election in Rhode Island, 1868 (en)
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  • 1868 (xsd:integer)
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  • no (en)
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  • Democratic Party (en)
  • National Union Party (en)
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  • 37.76
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  • United States presidential election in Rhode Island, 1860 (en)
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  • 1860 (xsd:integer)
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  • President (en)
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  • presidential (en)
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  • The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.Rhode Island voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%. (en)
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  • United States presidential election in Rhode Island, 1864 (en)
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