1852 United States presidential election in Iowa

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.

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%

President before election Millard Fillmore Whig Elected President Franklin Pierce Democratic

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

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 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 (other)	N/A	139	0.39%	0	0.00%
Total				35,364	100.00%	4	100.00%

References

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