Election in Arkansas
1836 United States presidential election in Arkansas
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| November 3 – December 7, 1836 | 1840 → |
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| |  |  | | Nominee | Martin Van Buren | Hugh White | | | Party | Democratic | Whig | | Home state | New York | Tennessee | | Running mate | Richard Johnson | John Tyler | | Electoral vote | 3 | 0 | | Popular vote | 2,380 | 1,334 | | Percentage | 64.08% | 35.92% | |
 County Results | Van Buren 50-60% 60-70% 70-80% 80-90% 90-100% | White 50-60% 60-70% | No Vote: | |
President before election
Andrew Jackson
Democratic | Elected President
Martin Van Buren
Democratic | |
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- 1996
- 2000
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- 2016
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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 1836 United States presidential election in Arkansas took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas, having been admitted to the Union as the 25th state on June 15, 1836, voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White during its first presidential election. Van Buren won Arkansas by a margin of 28.16%.
Results
| 1836 United States presidential election in Arkansas[1] |
| Party | Candidate | Votes | Percentage | Electoral votes |
| | Democratic | Martin Van Buren | 2,380 | 64.08% | 3 |
| | Whig | Hugh White | 1,334 | 35.92% | 0 |
| Totals | 3,714 | 100.0% | 3 |
See also
References
- ^ "1836 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved August 4, 2012.
