1874 Rhode Island gubernatorial election
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The 1874 Rhode Island gubernatorial election was held on April 1, 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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April 1, 1874
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 County results Howard: 60â70% 80â90% >90% | |||||||||||||||||
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General election
On election day, April 1, 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 May 5, 1874.[2]
Results
| 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 | ||||