1832 United States presidential election in Tennessee explained

See main article: 1832 United States presidential election.

Election Name:1832 United States presidential election in Tennessee
Type:presidential
Ongoing:no
Previous Election:1828 United States presidential election in Tennessee
Previous Year:1828
Next Election:1836 United States presidential election in Tennessee
Next Year:1836
Election Date:November 2 – December 5, 1832
Image1:Andrew jackson head.jpg
Nominee1:Andrew Jackson
Party1:Democratic Party (United States)
Home State1:Tennessee
Running Mate1:Martin Van Buren
Electoral Vote1:15
Popular Vote1:28,078
Percentage1:95.42%
Nominee2:Henry Clay
Party2:National Republican Party (United States)
Home State2:Kentucky
Running Mate2:John Sergeant
Electoral Vote2:0
Popular Vote2:1,347
Percentage2:4.58%
Map Size:350px
Precinct
Before Election:Andrew Jackson
Before Party:Democratic Party (United States)
After Election:Andrew Jackson
After Party:Democratic Party (United States)

The 1832 United States presidential election in Tennessee took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.

Tennessee voted for the Democratic Party candidate, state native Andrew Jackson, over the National Republican candidate, Henry Clay. Jackson won Tennessee by a large margin of 90.84%. This is the most recent election in which Blount County, Carter County, Cocke County, Grainger County, Jefferson County, and Sevier County voted for a Democratic presidential candidate on a presidential level.

Results

1832 United States presidential election in Tennessee[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticAndrew Jackson (incumbent)28,07895.42%15
width: 3px" National RepublicanHenry Clay1,3474.58%0
Totals29,425100.00%15

Notes and References

  1. Web site: 1832 Presidential General Election Results — Tennessee. U.S. Election Atlas. 12 April 2013.