1832 United States presidential election in Kentucky explained

See main article: 1832 United States presidential election.

Election Name:1832 United States presidential election in Kentucky
Type:presidential
Ongoing:no
Previous Election:1828 United States presidential election in Kentucky
Previous Year:1828
Next Election:1836 United States presidential election in Kentucky
Next Year:1836
Election Date:November 2 – December 5, 1832
Image1:Henry Clay (copy after Edward Dalton Marchant).jpg
Nominee1:Henry Clay
Party1:National Republican Party (United States)
Home State1:Kentucky
Running Mate1:John Sergeant
Electoral Vote1:15
Popular Vote1:43,449
Percentage1:54.49%
Nominee2:Andrew Jackson
Party2:Democratic Party (United States)
Home State2:Tennessee
Running Mate2:Martin Van Buren
Electoral Vote2:0
Popular Vote2:36,292
Percentage2:45.51%

The 1832 United States presidential election in Kentucky 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.

Kentucky voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Kentucky by a margin of 8.98%. His victory was likely influenced by the home state advantage he received from his native state. Kentucky and South Carolina were the only 2 states Jackson lost in 1832 that he had won in the previous cycle.

Results

United States presidential election in Kentucky, 1832[1]
PartyCandidateVotesPercentageElectoral votes
width: 3px" National RepublicanHenry Clay43,44954.49%15
DemocraticAndrew Jackson (incumbent)36,29245.51%0
Totals79,741100.0%15

Notes and References

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