1875 Rhode Island gubernatorial election

Rhode Island gubernatorial election
1875 Rhode Island gubernatorial election

← 1874 7 April 1875 1876 →
 
Nominee Henry Lippitt Rowland Hazard Charles R. Cutler
Party Republican Independent Republican Democratic
Alliance Prohibition
Electoral vote 70 36 eliminated
Popular vote 8,368 8,724 5,166
Percentage 37.6% 39.2% 23.2%

Governor before election

Henry Howard
Republican

Elected Governor

Henry Lippitt
Republican

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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

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

Rhode Island gubernatorial election, 1875[4]
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

  1. ^ "Henry Lippitt". National Governors Association. Retrieved 8 April 2024.
  2. ^ "Meeting of the General Assembly". Daily Kennebec Journal. May 26, 1875.
  3. ^ "Election of State Officers in Rhode Island". Worcester Daily Spy. May 26, 1875.
  4. ^ Gubernatorial Elections, 1787-1997. Washington, D. C.: Congressional Quarterly. 1998. p. 77.