1875 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%

County results Lippitt:      30–40% Hazard:      40–50% Cutler:      30–40%      40–50%

Governor before election Henry Howard Republican Elected Governor Henry Lippitt Republican

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v t e

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]