1869 Iowa gubernatorial election

1869 Iowa gubernatorial election

←  1867 October 12, 1869 1871  →

Nominee Samuel Merrill George Gillespie Party Republican Democratic Popular vote 97,243 57,287 Percentage 62.93% 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

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

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