1828 United States presidential election in Rhode Island

Main article: 1828 United States presidential election

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.

1828 United States presidential election in Rhode Island

← 1824 October 31 – December 2, 1828 1832 →

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%

County Results Adams   60-70%   70-80%   80-90%

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

• United States presidential elections in Rhode Island

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.