United States presidential election in Rhode Island, 1832

United States presidential election in Rhode Island, 1832

November 2 – December 5, 1832

Nominee Henry Clay Andrew Jackson Party National Republican Democratic Home state Kentucky Tennessee Running mate John Sergeant Martin Van Buren Electoral vote 4 0 Popular vote 2,810 2,126 Percentage 56.93% 43.07%

The 1832 United States presidential election in Rhode Island took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for the National Republican candidate, Henry Clay, over the Democratic Party candidate, Andrew Jackson. Clay won Rhode Island by a margin of 13.86%.

Results

United States presidential election in Rhode Island, 1832[1]
Party		Candidate	Votes	Percentage	Electoral votes
	National Republican	Henry Clay	2,810	56.93%	21
	Democratic	Andrew Jackson	2,126	43.07%	0
Totals			4,936	100.0%	4

References

1. "1832 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 12 April 2013.