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]
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County results Burnside: 60–70% 80–90% | |||||||||||||||||
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General election
editOn 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.
Results
editParty | Candidate | Votes | % | |
---|---|---|---|---|
Republican | Ambrose Burnside (incumbent) | 10,038 | 63.61 | |
Democratic | Lyman Pierce | 5,731 | 36.32 | |
Scattering | 11 | 0.07 | ||
Total votes | 15,780 | 100.00 | ||
Republican hold |
References
edit- ^ "Ambrose Burnside". National Governors Association. Retrieved 7 April 2024.
- ^ United States Gubernatorial Elections 1861-1911 (Michael J. Dubin) Page 478