1796 United States presidential election in New Jersey explained

See main article: 1796 United States presidential election.

Election Name:1796 United States presidential election in New Jersey
Country:New Jersey
Type:presidential
Ongoing:no
Previous Election:1792 United States presidential election in New Jersey
Previous Year:1792
Next Election:1800 United States presidential election in New Jersey
Next Year:1800
Election Date:November 4 - December 7, 1796
Image1:US Navy 031029-N-6236G-001 A painting of President John Adams (1735-1826), 2nd president of the United States, by Asher B. Durand (1767-1845)-crop.jpg
Nominee1:John Adams
Party1:Federalist Party (United States)
Home State1:Massachusetts
Electoral Vote1:7
Percentage1:100.00%
Nominee2:Thomas Pinckney
Party2:Federalist Party (United States)
Home State2:South Carolina
Electoral Vote2:7
Percentage2:
President
Before Election:George Washington
Before Party:Independent (politician)
After Election:John Adams
After Party:Federalist Party (United States)

The 1796 United States presidential election in New Jersey took place between November 4 and December 7, 1796, as part of the 1796 United States presidential election. The state legislature chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, New Jersey cast seven electoral votes for incumbent Vice President John Adams.[1]

See also

Notes and References

  1. Book: Dubin, Michael J. . United States Presidential Elections, 1788-1860: The Official Results by County and State . McFarland & Company . 2002 . 9780786410170 . Jefferson . 6.