1852 United States presidential election in Iowa

Main article: 1852 United States presidential election

1852 United States presidential election in Iowa

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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 17,763 15,856 Percentage 50.23% 44.84%

County Results Pierce   40–50%   50–60%   60–70%   70–80%   80–90% Scott   40–50%   50–60% Hale   40–50% Tie   ~50% Pierce & Scott

President before election Millard Fillmore Whig Elected President Franklin Pierce Democratic

The 1852 United States presidential election in Iowa 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.

Iowa voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Iowa by a margin of 5.39%.

This would be the last time Iowa would back a Democratic presidential nominee until 1912, and the last time it would be with an absolute majority of the vote until 1932.

Results

1852 United States presidential election in Iowa [1] [2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Franklin Pierce of New Hampshire	William R. King of Alabama	17,763	50.23%	4	100.00%
	Whig	Winfield Scott of New Jersey	William Alexander Graham of North Carolina	15,856	44.84%	0	0.00%
	Free Soil	John P. Hale of New Hampshire	George W. Julian of Indiana	1,606	4.54%	0	0.00%
		Write-in	N/A	139	0.39%	0	0.00%
Total				35,364	100.00%	4	100.00%

See also

• United States presidential elections in Iowa

References

1. "1852 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved December 1, 2017.
2. "1852 Presidential Election". The American Presidency Project. University of California Santa Barbara. Retrieved December 1, 2017.