| Election Name: | 1865 Iowa gubernatorial election |
| Country: | Iowa |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1863 Iowa gubernatorial election |
| Previous Year: | 1863 |
| Next Election: | 1867 Iowa gubernatorial election |
| Next Year: | 1867 |
| Election Date: | 10 October 1865 |
| Nominee1: | William M. Stone |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 70,461 |
| Percentage1: | 56.41% |
| Nominee2: | Thomas H. Benton |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 54,090 |
| Percentage2: | 43.31% |
| Map Size: | 240px |
| Governor | |
| Before Election: | William M. Stone |
| Before Party: | Republican Party (United States) |
| After Election: | William M. Stone |
| After Party: | Republican Party (United States) |
The 1865 Iowa gubernatorial election was held on 10 October 1865 in order to elect the Governor of Iowa. Incumbent Republican Governor William M. Stone was re-elected against Democratic nominee Thomas H. Benton.[1]
On election day, 10 October 1865, incumbent Republican Governor William M. Stone won re-election by a margin of 16,371 votes against his opponent Democratic nominee Thomas H. Benton, thereby holding Republican control over the office of Governor. Stone was sworn in for his second term on 16 January 1866.[2]