1868 Rhode Island gubernatorial election
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Elections in Rhode Island |
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The 1868 Rhode Island gubernatorial election was held on 1 April 1868 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 1866 and 1867 election.[1]
General election[edit]
On election day, 1 April 1868, incumbent Republican Governor Ambrose Burnside won re-election by a margin of 6,648 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of Governor. Burnside was sworn in for his third term on 4 May 1868.[2]
Results[edit]
Party | Candidate | Votes | % | |
---|---|---|---|---|
Republican | Ambrose Burnside (incumbent) | 10,038 | 74.69 | |
Democratic | Lyman Pierce | 3,390 | 25.23 | |
Scattering | 11 | 0.08 | ||
Total votes | 13,439 | 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:
- 1868 Rhode Island elections
- Rhode Island gubernatorial elections
- 1868 in Rhode Island
- 1868 United States gubernatorial elections
- April 1868 events
- 1860s in Rhode Island
- 1860s Rhode Island elections
- 1868 elections
- 1868 elections in North America
- 1868 elections in the United States
- Government of Rhode Island
- United States gubernatorial elections in the 1860s