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 County Results Lincoln 60–70%
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The1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the1864 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.
Rhode Island voted for theNational Union candidate, incumbentRepublican Party PresidentAbraham Lincoln and his running mateAndrew Johnson. They defeated theDemocratic candidate,George B. McClellan and his running mateGeorge H. Pendleton. Lincoln won the state by a margin of 24.48%.
| 1864 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| National Union | Abraham Lincoln (incumbent) | 13,962 | 62.24% | 4 | |
| Democratic | George B. McClellan | 8,470 | 37.76% | 0 | |
| Totals | 22,432 | 100.0% | 4 | ||
State results of the1864 U.S. presidential election | ||
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