1874 Rhode Island gubernatorial election
Appearance
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County results Howard: 60–70% 80–90% >90% | |||||||||||||||||
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Elections in Rhode Island |
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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]
General election
[edit]On 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
[edit]Party | Candidate | Votes | % | |
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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.
Categories:
- 1874 Rhode Island elections
- Rhode Island gubernatorial elections
- 1874 in Rhode Island
- 1874 United States gubernatorial elections
- April 1874 events
- 1870s in Rhode Island
- 1870s Rhode Island elections
- 1874 elections
- 1874 elections in North America
- 1874 elections in the United States
- Government of Rhode Island
- United States gubernatorial elections in the 1870s