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1832 United States presidential election in Rhode Island

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Main article:1832 United States presidential election
1832 United States presidential election in Rhode Island

← 1828
November 2 – December 5, 1832
1836 →
 
NomineeHenry ClayAndrew Jackson
PartyNational RepublicanDemocratic
Home stateKentuckyTennessee
Running mateJohn SergeantMartin Van Buren
Electoral vote40
Popular vote2,8102,126
Percentage56.93%43.07%

County Results

Clay

  40–50%
  50–60%

Jackson

  50–60%

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The1832 United States presidential election in Rhode Island took place between November 2 and December 5, 1832, as part of the1832 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forPresident andVice President.

Rhode Island voted for theNational Republican candidate,Henry Clay, over theDemocratic 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

[edit]
1832 United States presidential election in Rhode Island[1]
PartyCandidateVotesPercentageElectoral votes
National RepublicanHenry Clay2,81056.93%4
DemocraticAndrew Jackson (incumbent)2,12643.07%0
Totals4,936100.0%4

See also

[edit]

References

[edit]
  1. ^"1832 Presidential General Election Results - Rhode Island". U.S. Election Atlas. RetrievedApril 12, 2013.
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