1828 United States presidential election in Vermont

Main article: 1828 United States presidential election

The 1828 United States presidential election in Vermont took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

1828 United States presidential election in Vermont

← 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 7 0 Popular vote 25,363 8,350 Percentage 75.23% 24.77%

County Results Adams   50–60%   60–70%   70–80%   80–90%

Vermont voted for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Vermont by a margin of 50.46%.

With 75.23% of the popular vote, Adam's victory in Vermont was his third strongest state in the 1828 election after Rhode Island and Massachusetts.^[1] No Democrat was to win Vermont until Lyndon B. Johnson did so in 1964. This election marks the first time Vermont chose its electors by popular vote rather than appointment by the state legislature.^[2]

Results

1828 United States presidential election in Vermont[3]
Party		Candidate	Votes	Percentage	Electoral votes
	National Republican	John Quincy Adams (incumbent)	25,363	75.23%	7
	Democratic	Andrew Jackson	8,350	24.77%	0
Totals			33,713	100.0%	7

See also

• United States presidential elections in Vermont