United States presidential election in New Jersey, 1832

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United States presidential election in New Jersey, 1832

← 1828 November 2 – December 5, 1832 1836 →

  Andrew Jackson.jpg Henry Clay.JPG
Nominee Andrew Jackson Henry Clay
Party Democratic National Republican
Home state Tennessee Kentucky
Running mate Martin Van Buren John Sergeant
Electoral vote 8 0
Popular vote 23,826 23,466
Percentage 49.89% 49.13%

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

New Jersey voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won New Jersey by a margin of 0.76%.

Results

United States presidential election in New Jersey, 1832[1]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 23,826 49.89% 8
National Republican Henry Clay 23,466 49.13% 0
Anti-Masonic William Wirt 468 0.98% 0
Totals 47,760 100.0% 8

References

  1. ^ "1832 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 12 April 2013. 
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