See main article: 1836 United States presidential election.
| Election Name: | 1836 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1832 United States presidential election in Rhode Island |
| Previous Year: | 1832 |
| Next Election: | 1840 United States presidential election in Rhode Island |
| 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: | 4 |
| Popular Vote1: | 2,964 |
| Percentage1: | 52.24% |
| Nominee2: | William Henry Harrison |
| Party2: | Whig Party (United States) |
| Home State2: | Ohio |
| Running Mate2: | Francis Granger |
| Electoral Vote2: | 0 |
| Popular Vote2: | 2,710 |
| Percentage2: | 47.76% |
| Map Size: | 250px |
| President | |
| Before Election: | Andrew Jackson |
| Before Party: | Democratic Party (United States) |
| Before Color: | FF3333 |
| After Election: | Martin Van Buren |
| After Party: | Democratic Party (United States) |
| After Color: | FF3333 |
The 1836 United States presidential election in Rhode Island took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island voted for Democratic candidate Martin Van Buren over Whig candidate William Henry Harrison. Van Buren won Rhode Island by a narrow margin of 4.48%.
This was the first time that Rhode Island ever voted for a Democratic presidential candidate, and Van Buren's performance would not be bettered by a Democrat in Rhode Island until Franklin D. Roosevelt in 1932.[1]