The 1852 United States presidential election in Iowa took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
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 County Results
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Iowa voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Iowa by a margin of 5.39%.
This would be the last time Iowa would back a Democratic presidential nominee until 1912, and the last time it would be with an absolute majority of the vote until 1932.
Results
edit| 1852 United States presidential election in Iowa[1][2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Franklin Pierce of New Hampshire | William R. King of Alabama | 17,763 | 50.23% | 4 | 100.00% | ||
| Whig | Winfield Scott of New Jersey | William A. Graham of North Carolina | 15,856 | 44.84% | 0 | 0.00% | ||
| Free Soil | John P. Hale of New Hampshire | George W. Julian of Indiana | 1,606 | 4.54% | 0 | 0.00% | ||
| Write-in | N/A | 139 | 0.39% | 0 | 0.00% | |||
| Total | 35,364 | 100.00% | 4 | 100.00% | ||||
See also
editReferences
edit- ^ "1852 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved December 1, 2017.
- ^ "1852 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved December 1, 2017.
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