1875 Rhode Island gubernatorial election
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 County results Lippitt: 30–40% Hazard: 40–50% Cutler: 30–40% 40–50% | |||||||||||||||||||||||||||||
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| Elections in Rhode Island |
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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]
General election
[edit]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]
Results
[edit]| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Henry Lippitt | 8,368 | 37.6 | |
| Independent Republican | Rowland Hazard II | 8,724 | 39.2 | |
| Democratic | Charles R. Cutler | 5,166 | 23.2 | |
| Total votes | 22,258 | 100.00 | ||
| Republican hold | ||||
References
[edit]- ^ "Henry Lippitt". National Governors Association. Retrieved 8 April 2024.
- ^ "Meeting of the General Assembly". Daily Kennebec Journal. May 26, 1875.
- ^ "Election of State Officers in Rhode Island". Worcester Daily Spy. May 26, 1875.
- ^ Gubernatorial Elections, 1787-1997. Washington, D. C.: Congressional Quarterly. 1998. p. 77.
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