1796 United States presidential election in Rhode Island

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 4 representatives, or electors to the Electoral College who voted for president and vice president.

During this election, Rhode Island cast its 4 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

^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-08-29.
^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-08-29.
^ 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.

v t e State results of the 1796 U.S. presidential election
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Other 1796 elections	House Senate