Rhode Island · 1836 presidential election
Rhode Island presidential election results 1836
Martin Van Buren carried Rhode Island by 4.44 points (252 votes), winning 4 electoral votes.
- Electoral votes
- 4
- Votes counted
- 5,673
Statewide result
| Candidate | Party | Votes | Share | Bar | Electoral votes |
|---|---|---|---|---|---|
| Martin Van Buren | Democratic | 2,962 | 52.20% | 4 | |
| William Henry Harrison | Whig | 2,710 | 47.80% | — | |
| Hugh Lawson White | Whig | - | — | — | |
| Daniel Webster | Whig | — | — | — | |
| Total votes | 5,673 | ||||
Sources: 1836
Rhode Island in every election · All states in 1836 · Report an error
Questions and answers
Who won Rhode Island in 1836?
Martin Van Buren won Rhode Island by 4.44 percentage points (252 votes).
How many electoral votes did Rhode Island have in 1836?
Rhode Island cast 4 electoral votes in 1836.
How many people voted for president in Rhode Island in 1836?
5,673 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