1869 Iowa gubernatorial election
Appearance
| |||||||||||||||||
| |||||||||||||||||
|
Elections in Iowa |
---|
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
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
- ^ Kalb, Deborah (24 December 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved September 30, 2020.