1832 United States presidential election in Rhode Island

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1832 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1828 November 2 – December 5, 1832 1836  
  Henry Clay (copy after Edward Dalton Marchant).jpg Andrew jackson head.jpg
Nominee Henry Clay Andrew Jackson
Party National Republican Democratic
Home state Kentucky Tennessee
Running mate John Sergeant Martin Van Buren
Electoral vote40
Popular vote2,8102,126
Percentage56.93%43.07%

Rhode Island Presidential Election Results 1832.svg
County Results

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.

Contents

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

1832 United States presidential election in Rhode Island [1]
PartyCandidateVotesPercentageElectoral votes
National Republican Henry Clay 2,81056.93%4
Democratic Andrew Jackson (incumbent)2,12643.07%0
Totals4,936100.0%4

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References

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