1869 Rhode Island gubernatorial election

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1869 Rhode Island gubernatorial election
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←  1868 April 7, 1869 1870  →
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Nominee Seth Padelford Lyman Pierce
Party Republican Democratic
Popular vote7,3593,390
Percentage68.46%31.54%
1869 Rhode Island gubernatorial election results map by county.svg
County results
Padelford:      60–70%     70–80%     80–90%
Governor before election

Ambrose Burnside
Republican

Elected Governor

Seth Padelford
Republican

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]

Contents

Candidates

Republican Party

Democratic Party

Election

Statewide

1869 Rhode Island gubernatorial election [1]
PartyCandidateVotes%
Republican Seth Padelford 7,35968.46
Democratic Lyman Pierce3,39031.54
Total votes10,749 100.00
Republican hold

References

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


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