Svetlana Velmar-Janković
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Elections in Arkansas |
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A presidential election was held in Arkansas on November 2, 1840 as part of the 1840 United States presidential election.[1] 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
1840 United States presidential election in Arkansas[2] | ||||||||
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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 | 6,679 | 56.42% | 3 | 100.00% | ||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 5,160 | 43.58% | 0 | 0.00% | ||
Total | 11,839 | 100.00% | 3 | 100.00% |
See also
References
- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved December 23, 2013.