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The1824 United States presidential election in Rhode Island took place between October 26 and December 2, 1824, as part of the1824 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forPresident andVice President.
During this election, theDemocratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency.Rhode Island voted forJohn Quincy Adams overWilliam H. Crawford,Henry Clay, andAndrew Jackson. Adams won Rhode Island by a margin of 82.94%.
| 1824 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | John Quincy Adams | 2,145 | 91.47% | 4 | |
| Democratic-Republican | William H. Crawford | 200 | 8.53% | 0 | |
| Totals | 2,345 | 100.0% | 4 | ||
State and district results of the1824 United States presidential election | ||
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