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1869 Iowa gubernatorial election

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1869 Iowa gubernatorial election

← 1867October 12, 18691871 →
 
NomineeSamuel MerrillGeorge Gillespie
PartyRepublicanDemocratic
Popular vote97,24357,287
Percentage62.93%37.07%

County results
Merrill:     50-60%     60-70%     70-80%     80-90%     90-100%
Gillespie:     50-60%     60-70%
No Data/Votes:     

Governor before election

Samuel Merrill
Republican

Elected Governor

Samuel Merrill
Republican

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The1869 Iowa gubernatorial election was held on October 12, 1869. IncumbentRepublicanSamuel Merrill defeatedDemocratic 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]
PartyCandidateVotes%±%
RepublicanSamuel Merrill (incumbent)97,24362.93%
DemocraticGeorge Gillespie57,28737.07%
Majority39,956
Turnout
RepublicanholdSwing

References

[edit]
  1. ^Kalb, Deborah (December 24, 2015).Guide to U.S. Elections.ISBN 9781483380353. RetrievedSeptember 30, 2020.
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