| Election Name: | 1875 Rhode Island gubernatorial election |
| Country: | Rhode Island |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1874 Rhode Island gubernatorial election |
| Previous Year: | 1874 |
| Next Election: | 1876 Rhode Island gubernatorial election |
| Next Year: | 1876 |
| Election Date: | 7 April 1875 |
| Nominee1: | Henry Lippitt |
| Party1: | Republican Party (United States) |
| Electoral Vote1: | 70 |
| Popular Vote1: | 8,368 |
| Percentage1: | 37.6% |
| Nominee2: | Rowland Hazard |
| Party2: | Independent Republican (United States) |
| Alliance2: | Prohibition Party |
| Electoral Vote2: | 36 |
| Popular Vote2: | 8,724 |
| Percentage2: | 39.2% |
| Nominee3: | Charles R. Cutler |
| Party3: | Democratic Party (United States) |
| Electoral Vote3: | eliminated |
| Popular Vote3: | 5,166 |
| Percentage3: | 23.2% |
| Map Size: | 225px |
| Governor | |
| Before Election: | Henry Howard |
| Before Party: | Republican Party (United States) |
| After Election: | Henry Lippitt |
| After Party: | Republican Party (United States) |
The 1875 Rhode Island gubernatorial election took place on April 7, 1875 to elect the governor of Rhode Island. No candidate won a majority of the votes cast, sending the election to the Rhode Island General Assembly, where Henry Lippitt defeated fellow Republican Rowland Hazard II and Democrat Charles R. Cutler.[1]
The temperance movement divided Rhode Island Republicans ahead of the 1875 state elections. Lippitt was nominated by the "regular" Republican organization, but faced opposition from supporters of the state's temperance law. Pro-temperance independent Republicans and members of the Prohibition Party supported Hazard. On election day, Hazard received more votes than either of his rivals, but less than a majority. As stipulated by the Rhode Island Constitution, the election went to the General Assembly, which met on May 25 and elected Lippitt with 70 votes to 36 for Hazard.[2] [3]