1852 United States presidential election in Iowa

From Wikipedia, the free encyclopedia
Jump to navigation Jump to search
United States presidential election in Iowa, 1852

← 1848 November 2, 1852 1856 →
  Franklin Pierce by Brady.jpg Winfield Scott by Fredricks, 1862.jpg
Nominee Franklin Pierce Winfield Scott
Party Democratic Whig
Home state New Hampshire New Jersey
Running mate William R. King William A. Graham
Electoral vote 4 0
Popular vote 17,763 15,856
Percentage 50.23% 44.84%

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.

Iowa voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Iowa by a margin of 5.39%.

Results[edit]

United States presidential election in Iowa, 1852[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 (other) N/A 139 0.39% 0 0.00%
Total 35,364 100.00% 4 100.00%

References[edit]

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