| Elections in Rhode Island |
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The 1869 Rhode Island gubernatorial election took place on April 7, 1869 in order to elect the governor of Rhode Island.[1] Republican candidate and incumbent governor Seth Padelford won his first one-year term as governor[2] against Democratic candidate Lyman Pierce.[3]
Candidates
Republican Party
- Seth Padelford, the Republican nominee had previously served in the Rhode Island House of Representatives for two years, ran and lost in the 1860 Rhode Island gubernatorial election as the Republican nominee, and was the Lieutenant Governor of Rhode Island from 1863 to 1865 under Governor James Y. Smith.[2]
Democratic Party
- Lyman Pierce was the Democratic nominee.[3]
Election
Statewide
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Seth Padelford | 7,359 | 68.46 | |
| Democratic | Lyman Pierce | 3,390 | 31.54 | |
| Total votes | 10,749 | 100.00 | ||
| Republican hold | ||||
References
- ^ a b Dubin, Michael J. (2014). United States Gubernatorial Elections, 1861-1911 | The Official Results by State and County. McFarland. p. 5. ISBN 9780786456468.
- ^ a b "Seth Padelford". National Governors Association. 1 January 2019. Retrieved 27 March 2024.
- ^ a b "Democratic Nominations". Newport Daily News. 25 March 1869. Retrieved 27 March 2024 – via Newspapers.com.
