Jump to content

1869 Rhode Island gubernatorial election

From Wikipedia, the free encyclopedia

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

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

[edit]
  • Lyman Pierce was the Democratic nominee.[3]

Election

[edit]

Statewide

[edit]
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

[edit]
  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.