1832 United States presidential election in Rhode Island

Main article: 1832 United States presidential election

1832 United States presidential election in Rhode Island

←  1828 November 2 – December 5, 1832 1836  →

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%

County Results Clay   40-50%   50-60% Jackson   50-60%

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%.

Jackson remains the only Democrat to win two consecutive terms without carrying Rhode Island either time.

Results

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

See also

• United States presidential elections in Rhode Island

References

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