United States presidential election in Rhode Island, 1828

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United States presidential election in Rhode Island, 1828
Rhode Island
1824 ←
October 31 – December 2, 1828
→ 1832

  JohnQAdams.png Andrew Jackson.jpg
Nominee John Quincy Adams Andrew Jackson
Party National Republican Democratic
Home state Massachusetts Tennessee
Running mate Richard Rush John C. Calhoun
Electoral vote 4 0
Popular vote 2,754 821
Percentage 77.03% 22.97%

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%.

Results[edit]

United States presidential election in Rhode Island, 1828[1]
Party Candidate Votes Percentage Electoral votes
National Republican John Quincy Adams 2,754 77.03% 4
Democratic Andrew Jackson 821 22.97% 0
Totals 3,575 100.0% 4

References[edit]

  1. ^ "1828 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 28 February 2013.