1869 Iowa gubernatorial election

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1869 Iowa gubernatorial election

← 1867 October 12, 1869 1871 →
  Samuelmerrill.jpg 3x4.svg
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

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[edit]

Candidates[edit]

  • Samuel Merrill, Republican
  • George Gillespie, Democratic

Results[edit]

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

References[edit]

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