1824 United States presidential election in Rhode Island
Jump to navigation
Jump to search
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
County Results
Adams 80-90% 90-100%
| ||||||||||||||||||||||||||
|
Elections in Rhode Island |
---|
The 1824 United States presidential election in Rhode Island took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Rhode Island voted for John Quincy Adams over William H. Crawford, Henry Clay, and Andrew Jackson. Adams won Rhode Island by a margin of 82.94%.
Results
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 |
See also
References
- ^ "1824 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved February 27, 2013.
Categories:
- Articles with short description
- Short description matches Wikidata
- Use mdy dates from September 2023
- Elections using electoral votes
- 1824 United States presidential election by state
- United States presidential elections in Rhode Island
- 1824 Rhode Island elections
- All stub articles
- Rhode Island election stubs