Iowa · 1860 presidential election
Iowa presidential election results 1860
Abraham Lincoln carried Iowa by 11.92 points (15,298 votes), winning 4 electoral votes.
- Electoral votes
- 4
- Votes counted
- 128,331
- IA result
- R+11.9
- Compared with the nation
- 1.5 pts more Republican
- Swing since 1856
- 3.8 pts to R
Statewide result
| Candidate | Party | Votes | Share | Bar | Electoral votes |
|---|---|---|---|---|---|
| Abraham Lincoln | Republican | 70,409 | 54.87% | 4 | |
| Stephen Douglas | Democratic | 55,111 | 42.94% | — | |
| John Bell | Constitutional Union | 1,763 | 1.37% | — | |
| John C. Breckinridge | Southern Democratic | 1,048 | 0.82% | — | |
| Total votes | 128,331 | ||||
Sources: 1860
Iowa in every election · All states in 1860 · Report an error
Questions and answers
Who won Iowa in 1860?
Abraham Lincoln won Iowa by 11.92 percentage points (15,298 votes).
How many electoral votes did Iowa have in 1860?
Iowa cast 4 electoral votes in 1860.
How many people voted for president in Iowa in 1860?
128,331 votes were counted for president.
How did Iowa change from 1856 to 1860?
The margin moved 3.8 points toward the Republican ticket, from Republican by 8.1 points to Republican by 11.9 points.
Official results As published by The American Presidency Project (UC Santa Barbara) and the state documents it links. How we use this data