1869 Rhode Island gubernatorial election

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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%

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

Republican Party

  • 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

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

  1. ^ .
  2. ^ a b "Seth Padelford". National Governors Association. 1 January 2019. Retrieved 27 March 2024.
  3. ^
    Newspapers.com
    .