1832 United States presidential election in New Jersey

none June 2025 September 2023 1832 United States presidential election 1832 United States presidential election in New Jersey New Jersey presidential no 1828 United States presidential election in New Jersey 1828 1836 United States presidential election in New Jersey 1836 November 2 – December 5, 1832 x200px Andrew jackson head (cropped 4x5).jpg 180 150 200 0 20 Andrew Jackson Democratic Party (United States) Tennessee Martin Van Buren 8 23,826 49.89% Portrait of Henry Clay, 9th Secretary of State under President John Quincy Adams (Cropped1).jpg 180 150 200 0 15 Henry Clay National Republican Party (United States) Kentucky John Sergeant 0 23,466 49.13% File:1832 United States presidential election in New Jersey results map by county.svg 230px County results Jackson #86B6F2 50–60% #4389E3 60–70% #1666CB 70–80% Clay #FED463 50–60% #FE9929 60–70% 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%. This was the closest election in the state's history. Results 1832 United States presidential election in New Jersey1832 Presidential General Election Results - New Jersey http://uselectionatlas.org/RESULTS/state.php?year=1832&f=1&off=0&elect=0&fips=34&submit=Retrieve U.S. Election Atlas 12 April 2013PartyCandidateVotesPercentageElectoral votes Democratic Andrew Jackson (incumbent) 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 See also United States presidential elections in New Jersey References New Jersey 1832 Category:1832 New Jersey elections