1869 Iowa gubernatorial election

1869 Iowa gubernatorial election

← 1867 October 12, 1869 1871 →
 
Nominee Samuel Merrill George Gillespie
Party Republican Democratic
Popular vote 97,243 57,287
Percentage 62.93% 37.07%

Governor before election

Samuel Merrill
Republican

Elected Governor

Samuel Merrill
Republican

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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.

General election

Candidates

  • Samuel Merrill, Republican
  • George Gillespie, Democratic

Results

1869 Iowa gubernatorial election[1]
Party Candidate Votes % ±%
Republican Samuel Merrill (incumbent) 97,243 62.93%
Democratic George Gillespie 57,287 37.07%
Majority 39,956
Turnout
Republican hold Swing

References

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