See main article: 1900 United States presidential election.
| Election Name: | 1900 United States presidential election in Rhode Island |
| Country: | Rhode Island |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1896 United States presidential election in Rhode Island |
| Previous Year: | 1896 |
| Next Election: | 1904 United States presidential election in Rhode Island |
| Next Year: | 1904 |
| Election Date: | November 6, 1900 |
| Image1: | Mckinley (cropped).jpg |
| Nominee1: | William McKinley |
| Party1: | Republican Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | Theodore Roosevelt |
| Electoral Vote1: | 4 |
| Popular Vote1: | 33,784 |
| Percentage1: | 59.74% |
| Nominee2: | William Jennings Bryan |
| Party2: | Democratic Party (United States) |
| Home State2: | Nebraska |
| Running Mate2: | Adlai E. Stevenson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 19,812 |
| Percentage2: | 35.04% |
| Map Size: | 250px |
| President | |
| Before Election: | William McKinley |
| Before Party: | Republican Party (United States) |
| After Election: | William McKinley |
| After Party: | Republican Party (United States) |
The 1900 United States presidential election in Rhode Island took place on November 6, 1900, as part of the 1900 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won Rhode Island by a margin of 24.7% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.
Bryan had previous lost Rhode Island to McKinley four years earlier and would later lose the state again in 1908 to William Howard Taft.