1868 Rhode Island gubernatorial election
From Wikipedia, the free encyclopedia
The 1868 Rhode Island gubernatorial election was held on April 1, 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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April 1, 1868
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 County results Burnside: 60â70% 80â90% | |||||||||||||||||
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General election
On election day, April 1, 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 May 4, 1868.
Results
| Party | 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 | ||||