See main article: 1836 United States presidential election.
| Election Name: | 1836 United States presidential election in Arkansas |
| Country: | Arkansas |
| Type: | presidential |
| Ongoing: | no |
| Next Election: | 1840 United States presidential election in Arkansas |
| Next Year: | 1840 |
| Election Date: | November 3 – December 7, 1836 |
| Image1: | Martin Van Buren circa 1837 crop.jpg |
| Nominee1: | Martin Van Buren |
| Party1: | Democratic Party (United States) |
| Home State1: | New York |
| Running Mate1: | Richard Johnson |
| Electoral Vote1: | 3 |
| Popular Vote1: | 2,380 |
| Percentage1: | 64.08% |
| Nominee2: | Hugh White |
| Party2: | Whig Party (United States) |
| Home State2: | Tennessee |
| Running Mate2: | John Tyler |
| Electoral Vote2: | 0 |
| Popular Vote2: | 1,334 |
| Percentage2: | 35.92% |
| Map Size: | 250px |
| President | |
| Before Election: | Andrew Jackson |
| Before Party: | Democratic Party (United States) |
| After Election: | Martin Van Buren |
| After Party: | Democratic Party (United States) |
The 1836 United States presidential election in Arkansas took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas, having been admitted to the Union as the 25th state on June 15, 1836, voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White during its first presidential election. Van Buren won Arkansas by a margin of 28.16%.
| 1836 United States presidential election in Arkansas[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Martin Van Buren | 2,380 | 64.08% | 3 | |
| Whig | Hugh White | 1,334 | 35.92% | 0 | |
| Totals | 3,714 | 100.00% | 3 | ||