Rhode Island · 1828 presidential election
Rhode Island presidential election results 1828
John Quincy Adams carried Rhode Island by 54.05 points (1,935 votes), winning 4 electoral votes.
- Electoral votes
- 4
- Votes counted
- 3,580
Statewide result
| Candidate | Party | Votes | Share | Bar | Electoral votes |
|---|---|---|---|---|---|
| John Quincy Adams | National-Republican | 2,755 | 77.00% | 4 | |
| Andrew Jackson | Democratic | 820 | 22.90% | — | |
| Total votes | 3,580 | ||||
Sources: 1828
Rhode Island in every election · All states in 1828 · Report an error
Questions and answers
Who won Rhode Island in 1828?
John Quincy Adams won Rhode Island by 54.05 percentage points (1,935 votes).
How many electoral votes did Rhode Island have in 1828?
Rhode Island cast 4 electoral votes in 1828.
How many people voted for president in Rhode Island in 1828?
3,580 votes were counted for president.
Official results As published by The American Presidency Project (UC Santa Barbara) and the state documents it links. How we use this data