1796 United States presidential election in Rhode Island

1796 United States presidential election in Rhode Island

November 4 – December 7, 1796
 
Nominee John Adams
Party Federalist
Home state Massachusetts
Running mate Thomas Pinckney
Electoral vote 4
Percentage 100.00%

President before election

George Washington
Independent

Elected President

John Adams
Federalist

The 1796 United States presidential election in Rhode Island took place between November 4 to December 7, 1796, as part of the 1796 United States presidential election. Voters chose four representatives, or electors to the Electoral College who voted for president and vice president.

During this election, Rhode Island cast its four electoral votes for John Adams.[1]

Results

1796 United States presidential election in Rhode Island[2][3]: 6–8 
Party Candidate Votes Percentage Electoral votes
Independent George Washington (incumbent) 4
Federalist John Adams 4
Totals 8

See also

References

  1. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 29, 2024.
  2. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 29, 2024.
  3. ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788-1860: The Official Results by County and State. Jefferson, North Carolina: McFarland. p. xii. ISBN 0-7864-1017-5.