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| Elections in New Jersey |
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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. This is the last time that a Democrat won without carrying Middlesex County.
Results
| 1852 United States presidential election in New Jersey[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% | ||||
See also
References
- ^ "1852 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved December 24, 2013.
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