1866 Rhode Island gubernatorial election
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The 1866 Rhode Island gubernatorial election was held on April 4, 1866, in order to elect the governor of Rhode Island. Republican nominee and former Union Army Major General Ambrose Burnside defeated Democratic nominee Lyman Pierce.[1]
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April 4, 1866
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 County results Burnside: 60â70% 70â80% 80â90% | |||||||||||||||||
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General election
On election day, April 4, 1866, Republican nominee Ambrose Burnside won the election by a margin of 5,381 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Burnside was sworn in as the 30th governor of Rhode Island on May 1, 1866.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Ambrose Burnside | 8,197 | 73.36 | |
| Democratic | Lyman Pierce | 2,816 | 25.20 | |
| Scattering | 160 | 1.44 | ||
| Total votes | 11,221 | 100.00 | ||
| Republican hold | ||||