Election in Alabama
1840 United States presidential election in Alabama
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← 1836 | October 30 – December 2, 1840 | 1844 → |
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| | | Nominee | Martin Van Buren | William Henry Harrison | | Party | Democratic | Whig | Home state | New York | Ohio | Running mate | Richard Mentor Johnson | John Tyler | Electoral vote | 7 | 0 | Popular vote | 33,996 | 28,518 | Percentage | 54.38% | 45.62% | |
County Results Van Buren 50-60% 60-70% 70-80% 80-90% 90-100% | Harrison 50-60% 60-70% 70–80% | |
President before election Martin Van Buren Democratic | Elected President William Henry Harrison Whig | |
Elections in Alabama |
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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%. As of 2020, this remains the only time in American history that Alabama has voted for a different presidential candidate than neighboring Mississippi.
Results
See also
References
- ^ "1840 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved December 23, 2013.
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