1869 Iowa gubernatorial election
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 County results Merrill: 50-60% 60-70% 70-80% 80-90% 90-100% Gillespie: 50-60% 60-70% No Data/Votes: | |||||||||||||||||
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| Elections in Iowa |
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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
[edit]Candidates
[edit]- Samuel Merrill, Republican
- George Gillespie, Democratic
Results
[edit]| 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
[edit]- ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved September 30, 2020.