1828 United States presidential election in Rhode Island
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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 4 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
1828 United States presidential election in Rhode Island[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
National Republican | John Quincy Adams | 2,754 | 77.03% | 4 | |
Democratic | Andrew Jackson | 821 | 22.97% | 0 | |
Totals | 3,575 | 100.0% | 4 |
References
- ^ "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ^ "1828 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 28 February 2013.