1824 United States presidential election in Pennsylvania
Appearance
(Redirected from United States presidential election in Pennsylvania, 1824)
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County Results
Jackson 50-60% 60-70% 70-80% 80-90% 90-100%
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Elections in Pennsylvania |
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Government |
The 1824 United States presidential election in Pennsylvania took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 28 representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Pennsylvania voted for Andrew Jackson over John Quincy Adams, William H. Crawford, and Henry Clay. Jackson won Pennsylvania by a wide margin of 64.54%.
This is the only time any presidential candidate has swept every Pennsylvania county.
Results
[edit]1824 United States presidential election in Pennsylvania[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | Andrew Jackson | 35,929 | 76.04% | 28 | |
Democratic-Republican | John Quincy Adams | 5,436 | 11.50% | 0 | |
Democratic-Republican | William H. Crawford | 4,182 | 8.85% | 0 | |
Democratic-Republican | Henry Clay | 1,705 | 3.61% | 0 | |
Totals | 47,252 | 100.0% | 28 |
See also
[edit]References
[edit]- ^ "1824 Presidential General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved August 4, 2012.