See main article: 1836 United States presidential election.
| Election Name: | 1836 United States presidential election in Kentucky |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1832 United States presidential election in Kentucky |
| Previous Year: | 1832 |
| Next Election: | 1840 United States presidential election in Kentucky |
| Next Year: | 1840 |
| Election Date: | November 3 – December 7, 1836 |
| Image1: | William Henry Harrison by James Reid Lambdin, 1835 crop.jpg |
| Nominee1: | William Henry Harrison |
| Party1: | Whig Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | Francis Granger |
| Electoral Vote1: | 15 |
| Popular Vote1: | 36,861 |
| Percentage1: | 52.59% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | Richard Johnson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 33,229 |
| Percentage2: | 47.41% |
| Map Size: | 380px |
The 1836 United States presidential election in Kentucky took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
Kentucky voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Kentucky by a margin of 5.18%.
| United States presidential election in Kentucky, 1836[1] | |||||
|---|---|---|---|---|---|
| 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 | ||