| Election Name: | 1866 Rhode Island gubernatorial election |
| Country: | Rhode Island |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1865 Rhode Island gubernatorial election |
| Previous Year: | 1865 |
| Next Election: | 1867 Rhode Island gubernatorial election |
| Next Year: | 1867 |
| Election Date: | 4 April 1866 |
| Nominee1: | Ambrose Burnside |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 8,197 |
| Percentage1: | 73.36% |
| Nominee2: | Lyman Pierce |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 2,816 |
| Percentage2: | 25.20% |
| Map Size: | 250px |
| Governor | |
| Before Election: | James Y. Smith |
| Before Party: | Republican Party (United States) |
| After Election: | Ambrose Burnside |
| After Party: | Republican Party (United States) |
The 1866 Rhode Island gubernatorial election was held on 4 April 1866 in order to elect the Governor of Rhode Island. Republican nominee and former Union Army Major General Ambrose Burnside defeated Democratic nominee Lyman Pierce.[1]
On election day, 4 April 1866, Republican nominee Ambrose Burnside won the election by a margin of 5,381 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Burnside was sworn in as the 30th Governor of Rhode Island on 1 May 1866.[2]