1840 United States presidential election in Alabama
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.
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Elections in Alabama |
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Government |
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] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 33,996 | 54.38% | 7 | 100.00% | ||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 28,515 | 45.62% | 0 | 0.00% | ||
Total | 62,511 | 100.00% | 7 | 100.00% |
References
- "1840 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved December 23, 2013.
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