United States presidential election in Alabama, 1840

United States presidential election in Alabama, 1840

October 30 - December 2, 1840
Nominee Martin Van Buren William Henry Harrison Party Democratic Whig Home state New York Ohio Running mate none John Tyler Electoral vote 7 0 Popular vote 33,996 28,518 Percentage 54.38 45.62
President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

Elections in Alabama
Federal government Presidential elections 1820 1824 1828 1832 1836 1840 1844 1848 1852 1856 1860 1868 1872 1876 1880 1884 1888 1892 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1944 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 Presidential Primaries Democratic 2008 2016 Republican 2008 2012 U.S. Senate elections 1962 1980 1986 1990 1992 1996 1998 2002 2004 2008 2010 2014 2016 U.S. House elections 2000 2002 2004 2006 2008 2010 2012 2014 2016 Special elections 1st district, 2013 Senate, 2017
State government Gubernatorial elections 1894 1896 1898 1900 1902 1906 1910 1914 1918 1922 1926 1930 1934 1938 1942 1946 1950 1954 1958 1962 1966 1970 1974 1978 1982 1986 1990 1994 1998 2002 2006 2010 2014 State elections 2000 2002 2004 2006 2008 2010 2012 2014
Ballot measures 2006 Amendment 774
Mobile government Mayoral elections 2001 2005 2009 2013
Montgomery government Mayoral elections 2015
Government

The 1840 United States presidential election in Alabama took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Alabama voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Alabama by a margin of 8.76%.

Results

References