1853 Rhode Island gubernatorial election
The 1853 Rhode Island gubernatorial election was held on 6 April 1853 in order to elect the Governor of Rhode Island. Incumbent Democratic Governor Philip Allen won re-election against Whig nominee William W. Hoppin and Free Soil nominee Edward Harris.[1]
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General election
On election day, 6 April 1853, incumbent Democratic Governor Philip Allen won re-election by a margin of 2,243 votes against his foremost opponent William W. Hoppin, thereby keeping democratic control over the office of Governor. Free Soil nominee Edward Harris also was unable to be elected as Governor for the fourth time, after his election losses in the gubernatorial elections of 1849, 1850 and 1851. Allen was sworn in for his third term on 6 May 1853.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | Philip Allen (incumbent) | 10,471 | 54.40 | |
| Whig | William W. Hoppin | 8,228 | 42.75 | |
| Free Soil | Edward Harris | 542 | 2.82 | |
| Scattering | 7 | 0.03 | ||
| Total votes | 19,248 | 100.00 | ||
| Democratic hold | ||||
References
- Capace, Nancy (2001). The Encyclopedia of Rhode Island. North American Book Dist LLC. p. 180. ISBN 9780403096107.
- "RI Governor". ourcampaigns.com. June 6, 2005. Retrieved May 10, 2023.
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