1796 United States presidential election in New York explained

See main article: 1796 United States presidential election.

Election Name:1796 United States presidential election in New York
Country:New York
Flag Year:1778
Type:presidential
Ongoing:no
Previous Election:1792 United States presidential election in New York
Previous Year:1792
Next Election:1800 United States presidential election in New York
Next Year:1800
Election Date:November 4 – December 7, 1796
Image1:Official Presidential portrait of John Adams (by John Trumbull, circa 1792).jpg
Nominee1:John Adams
Party1:Federalist Party (United States)
Home State1:Massachusetts
Running Mate1:Thomas Pinckney
Electoral Vote1:12
Percentage1:100.00%
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 York took place between November 4 and December 7, 1796, as part of the 1796 United States presidential election. The state legislature chose 12 representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, New York cast 12 electoral votes for 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.