See main article: 1840 United States presidential election.
| Election Name: | 1840 United States presidential election in Arkansas |
| Country: | Arkansas |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1836 United States presidential election in Arkansas |
| Previous Year: | 1836 |
| Next Election: | 1844 United States presidential election in Arkansas |
| Next Year: | 1844 |
| Election Date: | October 30 – December 2, 1840 |
| Image1: | Martin Van Buren circa 1837 crop.jpg |
| Nominee1: | Martin Van Buren |
| Party1: | Democratic Party (United States) |
| Home State1: | New York |
| Running Mate1: | Richard M. Johnson |
| Electoral Vote1: | 3 |
| Popular Vote1: | 6,679 |
| Percentage1: | 56.42% |
| Nominee2: | William H. Harrison |
| Party2: | Whig Party (United States) |
| Home State2: | Ohio |
| Running Mate2: | John Tyler |
| Electoral Vote2: | 0 |
| Popular Vote2: | 5,160 |
| Percentage2: | 43.58% |
| President | |
| Before Election: | Martin Van Buren |
| Before Party: | Democratic Party (United States) |
| After Election: | William H. Harrison |
| After Party: | Whig Party (United States) |
The 1840 United States presidential election in Arkansas took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Arkansas by a margin of 12.84%.