1828 United States presidential election in Vermont
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 County Results
Adams 50-60% 60-70% 70-80% 80-90%
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| Elections in Vermont |
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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.
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[edit]
| 1828 United States presidential election in Vermont[3] | |||||
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| 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[edit]
References[edit]
- ^ "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ Moore, John L., ed. (1985). Congressional Quarterly's Guide to U.S. Elections (2nd ed.). Washington, D.C.: Congressional Quarterly, Inc. pp. 254–56.
- ^ "1828 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved February 28, 2013.
State and district results of the 1828 United States presidential election | ||
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