1828 United States presidential election in Ohio
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| Elections in Ohio |
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The 1828 United States presidential election in Ohio took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose 16 representatives, or electors to the Electoral College, who voted for President and Vice President.
Ohio voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won Ohio by a narrow margin of 3.2%.
Results
| 1828 United States presidential election in Ohio[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Andrew Jackson | 67,597 | 51.60% | 16 | |
| National Republican | John Quincy Adams | 63,396 | 48.40% | 0 | |
| Totals | 130,993 | 100.0% | 16 | ||
See also
References
- ^ "1828 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 28 February 2013.
State and district results of the 1828 United States presidential election | ||
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