| Election Name: | 1867 Rhode Island gubernatorial election |
| Country: | Rhode Island |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1866 Rhode Island gubernatorial election |
| Previous Year: | 1866 |
| Next Election: | 1868 Rhode Island gubernatorial election |
| Next Year: | 1868 |
| Election Date: | 3 April 1867 |
| Nominee1: | Ambrose Burnside |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 7,372 |
| Percentage1: | 69.84% |
| Nominee2: | Lyman Pierce |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 3,178 |
| Percentage2: | 30.11% |
| Map Size: | 250px |
| Governor | |
| Before Election: | Ambrose Burnside |
| Before Party: | Republican Party (United States) |
| After Election: | Ambrose Burnside |
| After Party: | Republican Party (United States) |
The 1867 Rhode Island gubernatorial election was held on 3 April 1867 in order to elect the Governor of Rhode Island. Incumbent Republican Governor Ambrose Burnside won re-election against Democratic nominee Lyman Pierce in a rematch of the previous election.[1]
On election day, 3 April 1867, incumbent Republican Governor Ambrose Burnside won re-election by a margin of 4,194 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Burnside was sworn in for his second term on 4 May 1867.[2]