See main article: 1840 United States presidential election.
| Election Name: | 1840 United States presidential election in Vermont |
| Country: | Vermont |
| Flag Year: | 1837 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1836 United States presidential election in Vermont |
| Previous Year: | 1836 |
| Next Election: | 1844 United States presidential election in Vermont |
| 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: | 7 |
| Popular Vote1: | 32,445 |
| Percentage1: | 63.90% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | none |
| Electoral Vote2: | 0 |
| Popular Vote2: | 18,009 |
| Percentage2: | 35.47% |
| Map Size: | 260px |
| President | |
| Before Election: | Martin Van Buren |
| Before Party: | Democratic Party (United States) |
| After Election: | William Henry Harrison |
| After Party: | Whig Party (United States) |
The 1840 United States presidential election in Vermont took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
Vermont voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 28.43%.
Harrison's 28.43% margin of victory made it his strongest victory in the election while he carried 63.90% of the popular vote made Vermont his second strongest state after Kentucky.[1]
Harrison had previously won Vermont against Van Buren four years earlier.