1840 United States presidential election in Alabama

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1840 United States presidential election in Alabama
Flag of Alabama.svg
  1836 October 30 – December 2, 1840 1844  
  Martin Van Buren circa 1837 crop.jpg William Henry Harrison crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Mentor Johnson John Tyler
Electoral vote70
Popular vote33,99628,518
Percentage54.38%45.62%

Alabama Presidential Election Results 1840.svg

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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.

Contents

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%. As of 2020, this remains the only time in American history that Alabama has voted for a different presidential candidate than neighboring Mississippi.

Results

1840 United States presidential election in Alabama [1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 33,99654.38%7100.00%
Whig William Henry Harrison of Ohio John Tyler of Virginia 28,51545.62%00.00%
Total62,511100.00%7100.00%

See also

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References

  1. "1840 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved December 23, 2013.