United States presidential election in Indiana, 1824

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United States presidential election in Indiana, 1824

← 1820 October 26 – December 2, 1824 1828 →
  Andrew Jackson.jpg Henry Clay.JPG JohnQAdams.png
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 vote 5 0 0
Popular vote 7,343 5,315 3,095
Percentage 46.61% 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.

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[edit]

United States presidential election in Indiana, 1824[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican Andrew Jackson 7,343 46.61% 5
Democratic-Republican Henry Clay 5,315 33.74% 0
Democratic-Republican John Quincy Adams 3,095 19.65% 0
Totals 15,753 100.0% 5

References[edit]

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