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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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 | % | |
|---|---|---|---|---|
| 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.