1828 United States presidential election in Rhode Island
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 County Results
Adams 60-70% 70-80% 80-90%
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| Elections in Rhode Island |
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The 1828 United States presidential election in Rhode Island took place between October 31 and December 2, 1828, as part of the 1828 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 Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Rhode Island by a margin of 54.06%.
With 77.03% of the popular vote, Adams' victory in Rhode Island made it his strongest state in the 1828 election.[1]
Results
[edit]| 1828 United States presidential election in Rhode Island[2] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| National Republican | John Quincy Adams (incumbent) | 2,754 | 77.03% | 4 | |
| Democratic | Andrew Jackson | 821 | 22.97% | 0 | |
| Totals | 3,575 | 100.0% | 4 | ||
See also
[edit]References
[edit]- ^ "1828 Presidential Election Statistics". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1828 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved February 28, 2013.
State and district results of the 1828 United States presidential election | ||
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