1852 United States presidential election in Iowa

1852 United States presidential election in Iowa

← 1848 November 2, 1852 1856 →

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

Elections in Iowa
Federal government Presidential elections 1848 1852 1856 1860 1864 1868 1872 1876 1880 1884 1888 1892 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020 Presidential caucuses Democratic 1976 1992 2000 2004 2008 2012 2016 2020 Republican 1996 2000 2008 2012 2016 2020 United States Senate elections 1918 1920 1922 sp 1924 1926 1926 sp 1930 1932 1936 1936 sp 1938 1942 1944 1948 1950 1954 1956 1960 1962 1966 1968 1972 1974 1978 1980 1984 1986 1990 1992 1996 1998 2002 2004 2008 2010 2014 2016 2020 2022 United States House elections 1960 1962 1964 1966 1968 1970 1972 1974 1976 1978 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000 2002 2004 2006 2008 2010 2012 2014 2016 2018 2020 2022
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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%.

Results[]

See also[]

• United States presidential elections in Iowa

References[]

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