1796 United States presidential election in Rhode Island

Main article: 1796 United States presidential election

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.

1796 United States presidential election in Rhode Island

← 1792 November 4 – December 7, 1796 1800 →

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

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

See also

• United States presidential elections in Rhode Island