Rhode Island · 1860 presidential election
Rhode Island presidential election results 1860
Abraham Lincoln carried Rhode Island by 22.44 points (4,487 votes), winning 4 electoral votes.
- Electoral votes
- 4
- Votes counted
- 19,993
- RI result
- R+22.4
- Compared with the nation
- 12.0 pts more Republican
- Swing since 1856
- 1.7 pts to D
Statewide result
| Candidate | Party | Votes | Share | Bar | Electoral votes |
|---|---|---|---|---|---|
| Abraham Lincoln | Republican | 12,240 | 61.22% | 4 | |
| Stephen Douglas | Democratic | 7,753 | 38.78% | — | |
| John C. Breckinridge | Southern Democratic | -- | — | — | |
| John Bell | Constitutional Union | -- | — | — | |
| Total votes | 19,993 | ||||
Sources: 1860
Rhode Island in every election · All states in 1860 · Report an error
Questions and answers
Who won Rhode Island in 1860?
Abraham Lincoln won Rhode Island by 22.44 percentage points (4,487 votes).
How many electoral votes did Rhode Island have in 1860?
Rhode Island cast 4 electoral votes in 1860.
How many people voted for president in Rhode Island in 1860?
19,993 votes were counted for president.
How did Rhode Island change from 1856 to 1860?
The margin moved 1.7 points toward the Democratic ticket, from Republican by 24.2 points to Republican by 22.4 points.
Official results As published by The American Presidency Project (UC Santa Barbara) and the state documents it links. How we use this data