1869 Iowa gubernatorial election

none January 2025 September 2023 1869 Iowa gubernatorial election Iowa Presidential no 1867 Iowa gubernatorial election 1867 1871 Iowa gubernatorial election 1871 October 12, 1869 x150px Samuel Merrill Republican Party (United States) 97,243 62.93% x150px George Gillespie Democratic Party (United States) 57,287 37.07% 1869 Iowa gubernatorial election results map by county.svg 240px County results Merrill: #e27f7f 50–60% #d75d5d 60–70% #D72F30 70–80% #C21B18 80–90% #A80000 90–100%Gillespie: #7996E2 50–60% #6674DE 60–70%No Data/Votes: #808080 Governor Samuel Merrill Republican Party (United States) Samuel Merrill Republican Party (United States) 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 electionhttps://books.google.com/books?id=9CZECwAAQBAJ&pg=PA1693 Guide to U.S. Elections 9781483380353 September 30, 2020 Kalb Deborah 24 December 2015 Republican Party (United States) Samuel Merrill (incumbent) 97,243 62.93% Democratic Party (United States) George Gillespie 57,287 37.07% 39,956 Republican Party (United States) References 1869 Iowa Category:1869 Iowa elections Category:October 1869 in the United States