Iowa · 1852 presidential election
Iowa presidential election results 1852
Franklin Pierce carried Iowa by 5.39 points (1,907 votes), winning 4 electoral votes.
- Electoral votes
- 4
- Votes counted
- 35,364
Statewide result
| Candidate | Party | Votes | Share | Bar | Electoral votes |
|---|---|---|---|---|---|
| Franklin Pierce | Democratic | 17,763 | 50.20% | 4 | |
| Winfield Scott | Whig | 15,856 | 44.80% | — | |
| John P. Hale | Free Soil | 1,606 | 4.50% | — | |
| Total votes | 35,364 | ||||
Sources: 1852
Iowa in every election · All states in 1852 · Report an error
Questions and answers
Who won Iowa in 1852?
Franklin Pierce won Iowa by 5.39 percentage points (1,907 votes).
How many electoral votes did Iowa have in 1852?
Iowa cast 4 electoral votes in 1852.
How many people voted for president in Iowa in 1852?
35,364 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