1840 United States presidential election in Tennessee

Main article: 1840 United States presidential election
1840 United States presidential election in Tennessee
November 3, 1840
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 15 0
Popular vote 60,194 47,951
Percentage 55.66% 44.34%
County results
President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

The 1840 United States presidential election in Tennessee was held on November 3, 1840 as part of the 1840 United States presidential election.[1] Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.

Tennessee voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won the state by a margin of 11.32%.

Results

1840 United States presidential election in Tennessee[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 60,194 55.66% 15 100.00%
Democratic Martin Van Buren of New York Richard Mentor Johnson of Kentucky 47,951 44.34% 0 0.00%
Total 108,145 100.00% 15 100.00%

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 - Tennessee". U.S. Election Atlas. Retrieved December 23, 2013.
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