How Can We Help?
You are here:
< Back

The 1828 United States presidential election in Rhode Island took place between October 31 and December 2, 1828, as part of the 1828 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, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Rhode Island by a margin of 54.06%.

With 77.03% of the popular vote, Adams' victory in Rhode Island made it his strongest state in the 1828 election.[1]

Results

1828 United States presidential election in Rhode Island[2]
Party Candidate Votes Percentage Electoral votes
National Republican John Quincy Adams (incumbent) 2,754 77.03% 4
Democratic Andrew Jackson 821 22.97% 0
Totals 3,575 100.0% 4

See also

References

  1. ^ "1828 Presidential Election Statistics". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1828 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved February 28, 2013.
Categories
Table of Contents