1852 United States presidential election in Iowa

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1852 United States presidential election in Iowa
Flag of Iowa.svg
  1848 November 2, 1852 1856  
  Mathew Brady - Franklin Pierce (cropped).jpg Winfield Scott by Fredricks, 1862 (cropped).jpg
Nominee Franklin Pierce Winfield Scott
Party Democratic Whig
Home state New Hampshire New Jersey
Running mate William R. King William A. Graham
Electoral vote40
Popular vote17,76315,856
Percentage50.23%44.84%

Iowa Presidential Election Results 1852.svg
County Results

President before election

Millard Fillmore
Whig

Elected President

Franklin Pierce
Democratic

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.

Contents

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

1852 United States presidential election in Iowa [1] [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Franklin Pierce of New Hampshire William R. King of Alabama 17,76350.23%4100.00%
Whig Winfield Scott of New Jersey William A. Graham of North Carolina 15,85644.84%00.00%
Free Soil John P. Hale of New Hampshire George W. Julian of Indiana 1,6064.54%00.00%
Write-inN/A1390.39%00.00%
Total35,364100.00%4100.00%

See also

References

  1. "1852 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved December 1, 2017.
  2. "1852 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved December 1, 2017.