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]
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![]() County results Howard: 60–70% 80–90% >90% | |||||||||||||||||
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General election
editOn 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]
Results
editParty | Candidate | Votes | % | |
---|---|---|---|---|
Republican | Henry Howard (incumbent) | 12,335 | 87.48 | |
Democratic | Lyman Pierce | 1,589 | 11.27 | |
Scattering | 177 | 1.25 | ||
Total votes | 14,101 | 100.00 | ||
Republican hold |
References
edit- ^ "Henry Howard". National Governors Association. Retrieved 8 April 2024.
- ^ "RI Governor". ourcampaigns.com. 6 October 2005. Retrieved 8 April 2024.