1824 United States presidential election in Indiana

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1824 United States presidential election in Indiana
Flag of Indiana.svg
  1820 October 26 – December 2, 1824 1828  
  Andrew Jackson.jpg Henry Clay.JPG John Quincy Adams 1858 crop.jpg
Nominee Andrew Jackson Henry Clay John Quincy Adams
Party Democratic-Republican Democratic-Republican Democratic-Republican
Home state Tennessee Kentucky Massachusetts
Running mate John C. Calhoun Nathan Sanford John C. Calhoun
Electoral vote500
Popular vote7,3435,3153,095
Percentage46.61%33.74%19.65%

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

The 1824 United States presidential election in Indiana took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

Contents

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Indiana voted for Andrew Jackson over Henry Clay and John Quincy Adams. Jackson won Indiana by a margin of 12.87%.

Results

1824 United States presidential election in Indiana [1]
PartyCandidateVotesPercentageElectoral votes
Democratic-Republican Andrew Jackson 7,34346.61%5
Democratic-Republican Henry Clay 5,31533.74%0
Democratic-Republican John Quincy Adams 3,09519.65%0
Totals15,753100.0%5

See also

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References

  1. "1824 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved February 27, 2013.