See main article: 1880 United States presidential election.
| Election Name: | 1880 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Flag Year: | 1877 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1876 United States presidential election in Rhode Island |
| Previous Year: | 1876 |
| Next Election: | 1884 United States presidential election in Rhode Island |
| Next Year: | 1884 |
| Election Date: | November 2, 1880 |
| Image1: | James Abram Garfield, photo portrait seated (cropped).jpg |
| Nominee1: | James A. Garfield |
| Party1: | Republican Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | Chester A. Arthur |
| Electoral Vote1: | 4 |
| Popular Vote1: | 18,195 |
| Percentage1: | 62.24% |
| Nominee2: | Winfield S. Hancock |
| Party2: | Democratic Party (United States) |
| Home State2: | Pennsylvania |
| Running Mate2: | William H. English |
| Electoral Vote2: | 0 |
| Popular Vote2: | 10,779 |
| Percentage2: | 36.87% |
| Map Size: | 250px |
| President | |
| Before Election: | Rutherford B. Hayes |
| Before Party: | Republican Party (United States) |
| After Election: | James A. Garfield |
| After Party: | Republican Party (United States) |
The 1880 United States presidential election in Rhode Island took place on November 2, 1880, as part of the 1880 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 nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state by a margin of 25.37%.
With 62.24% of the popular vote, Rhode Island would be Garfield's fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Minnesota.[1]