1832 United States presidential election in Rhode Island

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.

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

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

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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

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References

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