The 1867 Rhode Island gubernatorial election was held on 3 April 1867 in order to elect the governor of Rhode Island. Incumbent Republican governor Ambrose Burnside won re-election against Democratic nominee Lyman Pierce in a rematch of the previous election.[1]
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 County results Burnside: 60–70% 70–80% | |||||||||||||||||
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General election
editOn election day, 3 April 1867, incumbent Republican governor Ambrose Burnside won re-election by a margin of 4,194 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Burnside was sworn in for his second term on 4 May 1867.[2]
Results
edit| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Ambrose Burnside (incumbent) | 7,372 | 69.84 | |
| Democratic | Lyman Pierce | 3,178 | 30.11 | |
| Scattering | 6 | 0.05 | ||
| Total votes | 10,556 | 100.00 | ||
| Republican hold | ||||
References
edit- ^ "Ambrose Burnside". National Governors Association. Retrieved 7 April 2024.
- ^ "RI Governor". ourcampaigns.com. 26 July 2005. Retrieved 7 April 2024.