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 4 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 →
  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%

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%.

ResultsEdit

1832 United States presidential election in Rhode Island[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

ReferencesEdit

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