1852 United States presidential election in Iowa

November 2, 1852


Nominee	Franklin Pierce	Winfield Scott
Party	Democratic	Whig
Home state	New Hampshire	New Jersey
Running mate	William R. King	William A. 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

Elections in Iowa

Federal government

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Ballot measures

2020: Question 1

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Cedar Rapids

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Des Moines

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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
Party		Candidate	Running mate	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 A. 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%

References

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

v t e State results of the 1852 U.S. presidential election
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