United States presidential election in Rhode Island, 1832

From Infogalactic: the planetary knowledge core
Jump to: navigation, search

<templatestyles src="Module:Hatnote/styles.css"></templatestyles>

Main article: United States presidential election, 1832
United States presidential election in Rhode Island, 1832
← 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%

Lua error in package.lua at line 80: module 'strict' not found. 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

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

  1. ↑ Lua error in package.lua at line 80: module 'strict' not found.
Retrieved from "https://infogalactic.com/w/index.php?title=United_States_presidential_election_in_Rhode_Island,_1832&oldid=679112751"