1852 United States presidential election in Rhode Island

Main article: 1852 United States presidential election

The 1852 United States presidential election in Rhode Island took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

1852 United States presidential election in Rhode Island

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Nominee Franklin Pierce Winfield Scott Party Democratic Whig Home state New Hampshire New Jersey Running mate William R. King William Alexander Graham Electoral vote 4 0 Popular vote 8,735 7,626 Percentage 51.37% 44.85%

County Results Pierce   40–50%   50–60% Scott   50–60%   60–70%

President before election Millard Fillmore Whig Elected President Franklin Pierce Democratic

Rhode Island voted for the Democratic candidate, Franklin Pierce, over the Whig Party candidate, Winfield Scott. Pierce won the state by a margin of 6.52%. This was the first of three times that the state voted differently than Massachusetts (along with 1972 and 1980).

This would be the final time until 1912 that a Democratic presidential candidate was able to win Rhode Island and the final time until 1928 that a Democratic candidate won a majority of the popular vote.

Results

1852 United States presidential election in Rhode Island[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Franklin Pierce of New Hampshire	William Rufus DeVane King of Alabama	8,735	51.37%	4	100.00%
	Whig	Winfield Scott of New Jersey	William Alexander Graham of North Carolina	7,626	44.85%	0	0.00%
	Free Soil	John Parker Hale of New Hampshire	George Washington Julian of Indiana	644	3.79%	0	0.00%
Total				17,005	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island