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1856 United States presidential election in Iowa

United States presidential election in Iowa, 1856

← 1852 November 4, 1856 1860 →
  John Charles Fremont crop.jpg Hon. James Buchanan - NARA - 528318-crop.jpg Millard Fillmore by Brady Studio 1855-65-crop.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 Jackson Donelson
Electoral vote 4 0 0
Popular vote 45,073 37,568 9,669
Percentage 48.83% 40.70% 10.47%

President before election

Franklin Pierce

Elected President

James Buchanan

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.

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%.


United States presidential election in Iowa, 1856[1][2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican John C. Frémont of California William L. Dayton of New Jersey 45,073 48.83% 4 100.00%
Democratic James Buchanan of Pennsylvania John C. Breckinridge of Kentucky 37,568 40.70% 0 0.00%
Know Nothing Millard Fillmore of New York Andrew Jackson Donelson of Tennessee 9,669 10.47% 0 0.00%
Total 92,310 100.00% 4 100.00%


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