1869 Iowa gubernatorial election

1869 Iowa gubernatorial election
← 1867 October 12, 1869 1871 →
	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

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

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

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

[1]