1869 Rhode Island gubernatorial election

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]

1869 Rhode Island gubernatorial election
← 1868 April 7, 1869 1870 →
 
Nominee Seth Padelford Lyman Pierce
Party Republican Democratic
Popular vote 7,359 3,390
Percentage 68.46% 31.54%

County results
Padelford:      60–70%      70–80%      80–90%

Governor before election

Ambrose Burnside
Republican

Elected Governor

Seth Padelford
Republican

Candidates

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  • Lyman Pierce was the Democratic nominee.[3]

Election

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Statewide

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1869 Rhode Island gubernatorial election[1]
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

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  1. ^ a b Dubin, Michael J. (2014). United States Gubernatorial Elections, 1861-1911 | The Official Results by State and County. McFarland. p. 5. ISBN 9780786456468.
  2. ^ a b "Seth Padelford". National Governors Association. 1 January 2019. Retrieved 27 March 2024.
  3. ^ a b "Democratic Nominations". Newport Daily News. 25 March 1869. Retrieved 27 March 2024 – via Newspapers.com.