| Election Name: | 1873 Rhode Island gubernatorial election |
| Country: | Rhode Island |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1872 Rhode Island gubernatorial election |
| Previous Year: | 1872 |
| Next Election: | 1874 Rhode Island gubernatorial election |
| Next Year: | 1874 |
| Election Date: | 2 April 1873 |
| Nominee1: | Henry Howard |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 9,656 |
| Percentage1: | 71.68% |
| Nominee2: | Benjamin G. Chace |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 3,786 |
| Percentage2: | 28.11% |
| Map Size: | 225px |
| Governor | |
| Before Election: | Seth Padelford |
| Before Party: | Republican Party (United States) |
| After Election: | Henry Howard |
| After Party: | Republican Party (United States) |
The 1873 Rhode Island gubernatorial election was held on 2 April 1873 in order to elect the governor of Rhode Island. Republican nominee Henry Howard defeated Democratic nominee Benjamin G. Chace.[1]
On election day, 2 April 1873, Republican nominee Henry Howard won the election by a margin of 5,870 votes against his opponent Democratic nominee Benjamin G. Chace, thereby retaining Republican control over the office of governor. Howard was sworn in as the 32nd governor of Rhode Island on 6 May 1873.[2]