| Election Name: | 1869 Iowa gubernatorial election |
| Country: | Iowa |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1867 Iowa gubernatorial election |
| Previous Year: | 1867 |
| Next Election: | 1871 Iowa gubernatorial election |
| Next Year: | 1871 |
| Election Date: | October 12, 1869 |
| Nominee1: | Samuel Merrill |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 97,243 |
| Percentage1: | 62.93% |
| Nominee2: | George Gillespie |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 57,287 |
| Percentage2: | 37.07% |
| Map Size: | 240px |
| Governor | |
| Before Election: | Samuel Merrill |
| Before Party: | Republican Party (United States) |
| After Election: | Samuel Merrill |
| After Party: | 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.