1869 Iowa gubernatorial election

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1869 Iowa gubernatorial election
Flag of Iowa.svg
  1867 October 12, 1869 1871  
  Samuelmerrill.jpg 3x4.svg
Nominee Samuel Merrill George Gillespie
Party Republican Democratic
Popular vote97,24357,287
Percentage62.93%37.07%

Governor before election

Samuel Merrill
Republican

Elected Governor

Samuel Merrill
Republican

The 1869 Iowa gubernatorial election was held on October 12, 1869. Incumbent Republican Samuel Merrill defeated Democratic nominee George Gillespie with 62.93% of the vote.

Contents

General election

Candidates

Results

1869 Iowa gubernatorial election [1]
PartyCandidateVotes%±%
Republican Samuel Merrill (incumbent) 97,243 62.93%
Democratic George Gillespie57,28737.07%
Majority39,956
Turnout
Republican hold Swing

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References

  1. Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN   9781483380353 . Retrieved September 30, 2020.