1840 United States presidential election in Arkansas

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1840 United States presidential election in Arkansas
Flag of Arkansas.svg
  1836 November 2, 1840 1844  
  PA
CT  
  Martin Van Buren circa 1837 crop.jpg William Henry Harrison crop.jpg
Nominee Martin Van Buren William H. Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard M. Johnson John Tyler
Electoral vote30
Popular vote6,6795,160
Percentage56.42%43.58%

Arkansas Presidential Election Results 1840.svg

President before election

Martin Van Buren
Democratic

Elected President

William H. Harrison
Whig

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.

Contents

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]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 6,67956.42%3100.00%
Whig William Henry Harrison of Ohio John Tyler of Virginia 5,16043.58%00.00%
Total11,839100.00%3100.00%

See also

References

  1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
  2. "1840 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved December 23, 2013.