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.
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During this election, Rhode Island cast its 4 electoral votes for John Adams.[1]
Results
edit| 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
editReferences
edit- ^ "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.