See main article: 1828 United States presidential election.
| Election Name: | 1828 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1824 United States presidential election in Rhode Island |
| Previous Year: | 1824 |
| Next Election: | 1832 United States presidential election in Rhode Island |
| Next Year: | 1832 |
| Election Date: | October 31 – December 2, 1828 |
| Image1: | John Quincy Adams 1858 crop.jpg |
| Nominee1: | John Quincy Adams |
| Party1: | National Republican Party (United States) |
| Home State1: | Massachusetts |
| Running Mate1: | Richard Rush |
| Electoral Vote1: | 4 |
| Popular Vote1: | 2,754 |
| Percentage1: | 77.03% |
| Nominee2: | Andrew Jackson |
| Party2: | Democratic Party (United States) |
| Home State2: | Tennessee |
| Running Mate2: | John C. Calhoun |
| Electoral Vote2: | 0 |
| Popular Vote2: | 821 |
| Percentage2: | 22.97% |
| Map Size: | 250px |
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]
| 1828 United States presidential election in Rhode Island[2] | |||||
|---|---|---|---|---|---|
| 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 | ||