1856 United States presidential election in Rhode Island

Main article: 1856 United States presidential election

1856 United States presidential election in Rhode Island

November 4, 1856

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 11,467 6,680 1,675 Percentage 57.85% 33.70% 8.45%

County Results Frémont   50-60%   60-70%

President before election Franklin Pierce Democratic Elected President James Buchanan Democratic

The 1856 United States presidential election in Rhode Island 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.

Rhode Island voted for the Republican candidate, John C. Frémont, over the Democratic candidate, James Buchanan, and the Know Nothing candidate, Millard Fillmore. Frémont won the state by a margin of 24.15%.

With 57.85% of the popular vote, Rhode Island proved to be Frémont's fourth strongest state in the 1856 election after Vermont, Massachusetts and Maine.^[1]

Results

1856 United States presidential election in Rhode Island[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	John C. Frémont of California	William L. Dayton of New Jersey	11,467	57.85%	4	100.00%
	Democratic	James Buchanan of Pennsylvania	John C. Breckinridge of Kentucky	6,680	33.70%	0	0.00%
	Know Nothing	Millard Fillmore of New York	Andrew Jackson Donelson of Tennessee	1,675	8.45%	0	0.00%
Total				19,822	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island

References

1. "1856 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
2. "1856 Presidential General Election Results - Rhode Island".