The 1822 Rhode Island gubernatorial election was an uncontested election held on April 17, 1822 to elect the Governor of Rhode Island. William C. Gibbs, the Democratic-Republican nominee, was the only candidate and so won with 100% of the vote.
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General election edit
Candidates edit
- William C. Gibbs, Governor since 1821.[2]
Results edit
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Democratic-Republican | William C. Gibbs (incumbent) | 2,092 | 100% | ||
| Majority | 2,092 | 100% | |||
| Democratic-Republican hold | Swing | ||||
County results edit
| County[3] | William Gibbs
Democratic-Republican |
Total votes | |
|---|---|---|---|
| # | % | ||
| Bristol | 141 | 100% | 141 |
| Kent | 349 | 100% | 349 |
| Newport | 514 | 100% | 514 |
| Providence | 755 | 100% | 755 |
| Washington | 333 | 100% | 333 |
| Totals | 2,092 | 100% | 2,092 |
References edit
- ^ "Election". The Rhode-Island Republican. Newport, R.I. April 24, 1822. p. 2. Retrieved October 24, 2021.
- ^ "Governor William Channing Gibbs". National Governor's Association. Archived from the original on March 5, 2017.
- ^ a b "Rhode Island 1822 Governor". A New Nation Votes: American Election Returns 1787-1825. January 11, 2012. Retrieved June 2, 2022.
- ^ Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson: McFarland & Company. p. 232. ISBN 9780786414390.
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