1856 United States presidential election in Iowa

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.

1856 United States presidential election in Iowa

← 1852 November 4, 1856 1860 →

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 vote 4 0 0 Popular vote 45,073 37,568 9,669 Percentage 48.83% 40.70% 10.47%

County Results Frémont   40–50%   50–60%   60–70%   70–80%   80–90%   90–100% Buchanan   40–50%   50–60%   60–70%

President before election Franklin Pierce Democratic Elected President James Buchanan Democratic

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]
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%

See also

• United States presidential elections in Iowa