1867 Rhode Island gubernatorial election
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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]
General election[edit]
On 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.
Categories:
- 1867 Rhode Island elections
- Rhode Island gubernatorial elections
- 1867 in Rhode Island
- 1867 United States gubernatorial elections
- April 1867 events
- 1860s in Rhode Island
- 1860s Rhode Island elections
- 1867 elections
- 1867 elections in North America
- 1867 elections in the United States
- Government of Rhode Island
- United States gubernatorial elections in the 1860s