1852 United States presidential election in Alabama explained

See main article: 1852 United States presidential election.

Election Name:1852 United States presidential election in Alabama
Country:Alabama
Flag Year:1852
Type:presidential
Ongoing:no
Previous Election:1848 United States presidential election in Alabama
Previous Year:1848
Next Election:1856 United States presidential election in Alabama
Next Year:1856
Election Date:November 2, 1852
Image1:Mathew Brady - Franklin Pierce (cropped).jpg
Nominee1:Franklin Pierce
Party1:Democratic Party (United States)
Home State1:New Hampshire
Running Mate1:William R. King
Electoral Vote1:9
Popular Vote1:26,881
Percentage1:60.89%
Nominee2:Winfield Scott
Party2:Whig Party (United States)
Home State2:New Jersey
Running Mate2:William A. Graham
Electoral Vote2:0
Popular Vote2:15,061
Percentage2:34.12%
Image3:File:George M. Troup.jpg
Nominee3:George Troup
Party3:Southern Rights Party
Home State3:Alabama
Running Mate3:N/A
Electoral Vote3:0
Popular Vote3:2,205
Percentage3:5.01%
Map Size:200px
President
Before Election:Millard Fillmore
Before Party:Whig Party (United States)
After Election:Franklin Pierce
After Party:Democratic Party (United States)

The 1852 United States presidential election in Alabama took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.

Alabama voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Alabama by a margin of 26.77%.

See also