1856 United States presidential election in Georgia
The 1856 United States presidential election in Georgia took place on November 4, 1856, as part of the 1856 United States presidential election. Voters chose 10 representatives, or electors to the Electoral College, who voted for president and vice president.
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| Elections in Georgia |
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Georgia voted for the Democratic candidate, James Buchanan, over American Party candidate Millard Fillmore. Buchanan won Georgia by a margin of 14.28%.
Republican Party candidate John C. Frémont was not on the ballot in the state.
Results
| United States presidential election in Georgia, 1856[1][2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | James Buchanan of Pennsylvania | John C. Breckinridge of Kentucky | 56,581 | 57.14% | 10 | 100.00% | ||
| Know Nothing | Millard Fillmore of New York | Andrew Jackson Donelson of Tennessee | 42,439 | 42.86% | 0 | 0.00% | ||
| Total | 99,020 | 100.00% | 10 | 100.00% | ||||
References
- "1856 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved 3 December 2017.
- "1856 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved 3 December 2017.
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