1840 United States presidential election in Tennessee

Main article: 1840 United States presidential election

1840 United States presidential election in Tennessee

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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 Harrison   50–60%   60–70%   70–80%   80–90%   90–100% Van Buren   50–60%   60–70%   70–80%   80–90% Unknown/No vote

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 Tennessee by a margin of 11.32%.

Conventions

Main article: 1840 Whig National Convention

Main article: 1840 Democratic National Convention

Both William Henry Harrison and Martin Van Buren won their respective party conventions.

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 M. Johnson of Kentucky	47,951	44.34%	0	0.00%
Total				108,145	100.00%	15	100.00%

See also

• United States presidential elections in Tennessee

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.