Election in Arkansas
1840 United States presidential election in Arkansas
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| ← 1836 | October 30 – December 2, 1840 | 1844 → |
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| |  |  | | Nominee | Martin Van Buren | William H. Harrison | | | Party | Democratic | Whig | | Home state | New York | Ohio | | Running mate | Richard M. Johnson | John Tyler | | Electoral vote | 3 | 0 | | Popular vote | 6,679 | 5,160 | | Percentage | 56.42% | 43.58% | |
 | Van Buren 50-60% 60-70% 70-80% 80-90% 90-100% | Harrison 50-60% 60-70% 70-80% 80-90% | Unknown/No vote | |
President before election
Martin Van Buren
Democratic | Elected President
William H. Harrison
Whig | |
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| Mayoral elections |
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- 1996
- 2000
- 2004
- 2008
- 2012
- 2016
- 2020
- 2024
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| Mayoral elections |
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- 1996
- 2000
- 2004
- 2008
- 2012
- 2016
- 2020
- 2024
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| Mayoral elections |
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- 1996
- 2000
- 2004
- 2008
- 2012
- 2016
- 2020
- 2024
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The 1840 United States presidential election in Arkansas took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Arkansas by a margin of 12.84%.
Results
See also
References
- ^ "1840 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved December 23, 2013.
