See main article: 1840 United States presidential election.
| Election Name: | 1840 United States presidential election in Kentucky |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1836 United States presidential election in Kentucky |
| Previous Year: | 1836 |
| Next Election: | 1844 United States presidential election in Kentucky |
| Next Year: | 1844 |
| Election Date: | October 30 - December 2, 1840 |
| Image1: | William Henry Harrison crop.jpg |
| Nominee1: | William Henry Harrison |
| Party1: | Whig Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | John Tyler |
| Electoral Vote1: | 15 |
| Popular Vote1: | 58,488 |
| Percentage1: | 64.20% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | none |
| Electoral Vote2: | 0 |
| Popular Vote2: | 32,616 |
| Percentage2: | 35.80% |
| Map Size: | 380px |
| President | |
| Before Election: | Martin Van Buren |
| Before Party: | Democratic Party (United States) |
| Before Color: | FF3333 |
| After Election: | William Henry Harrison |
| After Party: | Whig Party (United States) |
| After Color: | FF3333 |
The 1840 United States presidential election in Kentucky took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
Kentucky voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Kentucky by a margin of 28.4%.
With 64.20% of the popular vote, Kentucky would prove to be Harrison's strongest state in the 1840 election.[1]