1856 United States presidential election in Iowa explained

See main article: 1856 United States presidential election.

Election Name:1856 United States presidential election in Iowa
Country:Iowa
Flag Year:1856
Type:presidential
Ongoing:no
Previous Election:1852 United States presidential election in Iowa
Previous Year:1852
Next Election:1860 United States presidential election in Iowa
Next Year:1860
Election Date:November 4, 1856
Image1:John Charles Fremont crop.jpg
Nominee1:John C. Frémont
Party1:Republican Party (United States)
Home State1:California
Running Mate1:William L. Dayton
Electoral Vote1:4
Popular Vote1:45,073
Percentage1:48.83%
Nominee2:James Buchanan
Party2:Democratic Party (United States)
Home State2:Pennsylvania
Running Mate2:John C. Breckinridge
Electoral Vote2:0
Popular Vote2:37,568
Percentage2:40.70%
Image3:Fillmore (cropped).jpg
Nominee3:Millard Fillmore
Party3:Know Nothing
Home State3:New York
Running Mate3:Andrew J. Donelson
Electoral Vote3:0
Popular Vote3:9,669
Percentage3:10.47%
Map Size:300px
President
Before Election:Franklin Pierce
Before Party:Democratic Party (United States)
After Election:James Buchanan
After Party:Democratic Party (United States)

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%.

Buchanan is the second of only 6 US presidents and the first of 4 Democratic presidents to have never won Iowa.

See also