1856 United States presidential election in Iowa

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1856 United States presidential election in Iowa
Flag of Iowa.svg
  1852 November 4, 1856 1860  
  John Charles Fremont crop.jpg James Buchanan (cropped).jpg Fillmore (cropped).jpg
Nominee John C. Frémont James Buchanan Millard Fillmore
Party Republican Democratic Know Nothing
Home state California Pennsylvania New York
Running mate William L. Dayton John C. Breckinridge Andrew J. Donelson
Electoral vote400
Popular vote45,07337,5689,669
Percentage48.83%40.70%10.47%

Iowa Presidential Election Results 1856.svg
County Results

President before election

Franklin Pierce
Democratic

Elected President

James Buchanan
Democratic

The 1856 United States presidential election in Iowa took place on November 4, 1856, as part of the 1856 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 Republican candidate, John C. Frémont, over Democratic candidate, James Buchanan and American Party candidate Millard Fillmore. Frémont won Iowa by a margin of 8.13%.

Buchanan is the second of only 6 US presidents and the first of 4 Democratic presidents to have never won Iowa. He also didn't carry Buchanan County, which is named after him.

Results

1856 United States presidential election in Iowa [1] [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Republican John C. Frémont of California William L. Dayton of New Jersey 45,07348.83%4100.00%
Democratic James Buchanan of Pennsylvania John C. Breckinridge of Kentucky 37,56840.70%00.00%
Know Nothing Millard Fillmore of New York Andrew Jackson Donelson of Tennessee 9,66910.47%00.00%
Total92,310100.00%4100.00%

See also

References

  1. "1856 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved December 3, 2017.
  2. "1856 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved December 3, 2017.