1856 United States presidential election in Iowa

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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.

Quick facts Nominee, Party ...
1856 United States presidential election in Iowa

← 1852
November 4, 1856
1860 â†’
 
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%

County Results

President before election

Franklin Pierce
Democratic

Elected President

James Buchanan
Democratic

Close

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

More information Party, Candidate ...
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%
Close

See also

References

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