1852 United States presidential election in New Jersey

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United States presidential election in New Jersey, 1852

← 1848 November 2, 1852 1856 →
  Franklin Pierce by Brady.jpg Winfield Scott by Fredricks, 1862.jpg
Nominee Franklin Pierce Winfield Scott
Party Democratic Whig
Home state New Hampshire New Jersey
Running mate William R. King William Alexander Graham
Electoral vote 7 0
Popular vote 44,305 38,556
Percentage 53.24% 44.97%

President before election

Millard Fillmore

Elected President

Franklin Pierce

The 1852 United States presidential election in New Jersey took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey voted for the Democratic candidate, Franklin Pierce, over the Whig Party candidate, Winfield Scott. Pierce won the state by a margin of 6.91%, making him the first Democratic presidential candidate since Andrew Jackson in 1832 to win the state.


United States presidential election in New Jersey, 1852[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Pierce of New Hampshire William Rufus DeVane King of Alabama 44,305 53.24% 7 100.00%
Whig Winfield Scott of New Jersey William Alexander Graham of North Carolina 38,556 46.33% 0 0.00%
Free Soil John Parker Hale of New Hampshire George Washington Julian of Indiana 359 0.43% 0 0.00%
Total 83,220 100.00% 7 100.00%


  1. ^ "1852 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 24 December 2013.

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