| Election Name: | 1874 Rhode Island gubernatorial election |
| Country: | Rhode Island |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1873 Rhode Island gubernatorial election |
| Previous Year: | 1873 |
| Next Election: | 1875 Rhode Island gubernatorial election |
| Next Year: | 1875 |
| Election Date: | 1 April 1874 |
| Nominee1: | Henry Howard |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 12,335 |
| Percentage1: | 87.48% |
| Nominee2: | Lyman Pierce |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 1,589 |
| Percentage2: | 11.27% |
| Map Size: | 225px |
| Governor | |
| Before Election: | Henry Howard |
| Before Party: | Republican Party (United States) |
| After Election: | Henry Howard |
| After Party: | Republican Party (United States) |
The 1874 Rhode Island gubernatorial election was held on 1 April 1874 in order to elect the Governor of Rhode Island. Incumbent Republican Governor Henry Howard won re-election against Democratic nominee Lyman Pierce.[1]
On election day, 1 April 1874, incumbent Republican Governor Henry Howard won re-election by a margin of 10,746 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Howard was sworn in for his second term on 5 May 1874.[2]