1852 United States presidential election in Iowa
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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| Elections in Iowa |
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State government |
Iowa voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Iowa by a margin of 5.39%.
Results
| 1852 United States presidential election in Iowa[1][2] | ||||||||
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| 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 (other) | N/A | 139 | 0.39% | 0 | 0.00% | |||
| Total | 35,364 | 100.00% | 4 | 100.00% | ||||
References
- "1852 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 1 December 2017.
- "1852 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved 1 December 2017.
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