United States presidential election in Rhode Island, 1832

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United States presidential election in Rhode Island, 1832
Rhode Island
1828 ←
November 2 – December 5, 1832 → 1836

  Henry Clay.JPG Andrew Jackson.jpg
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[edit]

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[edit]

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