1836 United States presidential election in Kentucky
From Wikipedia, the free encyclopedia
The 1836 United States presidential election in Kentucky was held between November 7 and 9, 1836 as part of the 1836 United States presidential election.[1] Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
November 7â9, 1836
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
County Results
| ||||||||||||||||||||||||||
Kentucky voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Kentucky by a margin of 5.18%.
Results
| 1836 United States presidential election in Kentucky[2] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Whig | William Henry Harrison | 36,861 | 52.59% | 15 | |
| Democratic | Martin Van Buren | 33,229 | 47.41% | 0 | |
| Totals | 70,090 | 100.0% | 15 | ||