See main article: 1856 United States presidential election.
| Election Name: | 1856 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1852 United States presidential election in Rhode Island |
| Previous Year: | 1852 |
| Next Election: | 1860 United States presidential election in Rhode Island |
| Next Year: | 1860 |
| Election Date: | November 4, 1856 |
| Image1: | John Charles Fremont crop.jpg |
| Nominee1: | John C. Frémont |
| Party1: | Republican Party (United States) |
| Home State1: | California |
| Running Mate1: | William L. Dayton |
| Electoral Vote1: | 4 |
| Popular Vote1: | 11,467 |
| Percentage1: | 57.85% |
| Nominee2: | James Buchanan |
| Party2: | Democratic Party (United States) |
| Home State2: | Pennsylvania |
| Running Mate2: | John C. Breckinridge |
| Electoral Vote2: | 0 |
| Popular Vote2: | 6,680 |
| Percentage2: | 33.70% |
| Image3: | Fillmore (cropped).jpg |
| Nominee3: | Millard Fillmore |
| Party3: | Know Nothing |
| Home State3: | New York |
| Running Mate3: | Andrew J. Donelson |
| Electoral Vote3: | 0 |
| Popular Vote3: | 1,675 |
| Percentage3: | 8.45% |
| Map Size: | 250px |
| President | |
| Before Election: | Franklin Pierce |
| Before Party: | Democratic Party (United States) |
| After Election: | James Buchanan |
| After Party: | Democratic |
The 1856 United States presidential election in Rhode Island took place on November 4, 1856, as part of the 1856 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 the Republican candidate, John C. Frémont, over the Democratic candidate, James Buchanan, and the Know Nothing candidate, Millard Fillmore. Frémont won the state by a margin of 24.15%.
With 57.85% of the popular vote, Rhode Island proved to be Frémont's fourth strongest state in the 1856 election after Vermont, Massachusetts and Maine.[1]