1832 United States presidential election in Rhode Island

none June 2025 September 2023 1832 United States presidential election 1832 United States presidential election in Rhode Island Rhode Island presidential no 1828 United States presidential election in Rhode Island 1828 1836 United States presidential election in Rhode Island 1836 November 2 – December 5, 1832 x200px Portrait of Henry Clay, 9th Secretary of State under President John Quincy Adams.jpg Henry Clay National Republican Party (United States) Kentucky John Sergeant 4 2,810 56.93% Andrew jackson head.jpg Andrew Jackson Democratic Party (United States) Tennessee Martin Van Buren 0 2,126 43.07% 1832 United States presidential election in Rhode Island results map by county.svg 250px County results Clay #FFEBCB 40–50% #FFDCA3 50–60% Jackson #86B6F2 50–60% 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%. Jackson remains the only Democrat to win two consecutive terms without carrying Rhode Island either time. Results 1832 United States presidential election in Rhode Island1832 Presidential General Election Results - Rhode Island http://uselectionatlas.org/RESULTS/state.php?year=1832&f=1&off=0&elect=0&fips=44&submit=Retrieve U.S. Election Atlas 12 April 2013PartyCandidateVotesPercentageElectoral 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 United States presidential elections in Rhode Island References Rhode Island 1832 Category:1832 Rhode Island elections