1832 United States presidential election in Alabama

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1832 United States presidential election in Alabama

← 1828 November 2 – December 5, 1832 1836 →
  Andrew jackson head.jpg
Nominee Andrew Jackson
Party Democratic
Home state Tennessee
Running mate Martin Van Buren
Electoral vote 7
Popular vote 14,286
Percentage 99.97%

President before election

John Quincy Adams
Democratic-Republican

Elected President

Andrew Jackson
Democratic

The 1832 United States presidential election in Alabama took place between November 2 and December 5, 1832, as part of the 1832 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, Andrew Jackson, over the National Republican candidate, Henry Clay. Jackson won Alabama by a margin of 99.94%.

Results[]

1832 United States presidential election in Alabama[1]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 14,286 99.97% 7
National Republican Henry Clay 5 0.03% 0
Totals 14,291 100.0% 7

References[]

  1. ^ "1832 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved April 10, 2013.


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